MCAT Physics Review

• B is correct. Watts can be written as J/s. Joules can be broken down further to N*m, which can be broken down further to kg*m/s².
• Thus:  W=J/s=
• (N*m)/s=
• (kg(m/s²)m)/s=
• kg*m²/s³

• 360° =  radians = 6.28 radians
• (so 180° =  radians = 3.14 radians)

• sin 30° = 0.50
• sin 45° = 0.70
• sin 60° = 0.87
• sin 90° = 1

• cos 30° = 0.87
• cos 45° = 0.70
• cos 60° = 0.50
• cos 90° = 0

tan 45° = 1

• 
• The multiple represents the hypotenuse.
• The  multiple represents the side opposite the 60° angle.
•  represents the side opposite the 30° angle.

• 
• The  multiple is associated with the hypotenuse.
•  is the length of each leg.

5:3 The right triangle would have the ratio of 3:4:5 with an angle approximately 53° between the 3 and 5 sides, and 37° between the 4 and 4 sides.

• The triangle conforms to the right triangle with a ratio of 1:2 (sides 4 and 8).
• 4(1)=4     4(2)=8
• The triangle with sides in the ratio of always has angles of 30°, 60°, 90°

• Given that A=8m and the cos30° =  approximately = 8.87, so A_x = 8*(0.87) = 7m.
• Similarly: sin30°=A_y/A and since sin30°=0.5, A_y = Asin30°= 8*(0.50)=4m

•  (units: m/s)
• where  is the change in time
• If there is zero displacement, the average velocity will also be zero.

Speed is the magnitude of velocity.

a=  = (units: m/s²)

• 
• v_f^o=0²+2(5m/s²)(40m) =√400 =20m/s

d is proportional to t² so since T doubles, T² will quadruple and therefore so will D. The answer is 4D.

• v_f is proportional to t
• d is proportional to t²
• v_f² is proportional to d

• Using the subscript "y" to distinguish from the x-direction...
• 
• 
• 

• 
• 80m=0+1/2(10m/s²)t²
• t=4s
• 
• v_fy=0+(10m/s²)(4s)
• v_fy=40m/s

• d_y=v_oyt+1/2(at2)
• d_y=(20m/s)(6s)+1/2(-10m/s2)(6s)²
• d_y=120m-180m = -60m
• d_y is negative only because upward was chosen as positive. Therefore h=60m.

• 1. When an object moves upward, it slows down by approximately 10 m/s every second.
• 2. When an object moves downward, it speeds up by approximately 10 m/s every second.
• 3. An object projected upward from the ground will take the same amount of time to travel up as it will to travel back to the ground.
• 4. An object projected upward from the ground will land with the same speed as it was initially projected.

• a) Since an object slows down 10m/s every second on the way up, it will take 5s to slow down from 50m/s to 0. Thus, t=5s.
• b) Since time up equals time down, the total time is 10s.
• c) Since the object will land with the same speed as initially projected, the final speed = 50m/s.

Since horizontal velocity never changes, it is equal to: =(40m/s)(0.5)=20m/s

• v_fy=v_oy+at
• d_y=v_oyt+(1/2)at²
• v_fy²=v_oy²+2ad_y

d_x=v_oxt

• First find t using the y- equations:
• 20m=0+(1/2)(10m/s²)t²
• t= 2s

• then plug into the x- equation:
• d_x=(15m/s)(2s) = 30m

• (a) To find the time up, look at your y- equations: 0=50m/s +(-10m/s²)t
• t=5s
• (b) To find height, use y- equation for d: d_y=(50m/s)(5s)+(1/2)(-10m/s²)5s)² = 250m-125m = 125m
• (c)Now use your x- equation: d_x=(87m/s)(10s)=870m

 where v=speed and r=radius

 where v= speed and T=period or time required for one complete revolution

a_c= (15m/s)²/30m = 7.5 m/s²

T=(2(3.14)(30m))/15m/s= approx. 12s

Since a_c=v²/r and r is a constant, we see that a_c is proportional to v², so it v doubles, then a_c would need to quadruple to maintain circular motion.

Inertia: An object initially at rest of in motion with a constant velocity will remain in its initial state unless acted upon by a nonzero net external force.

F_(net)=ma (net applied force equals the object's mass multiplied by the acceleration produced. It is in kilogram (kg) x m/s² or newtons (N0.

Sum of applied forces: F_net= F₁ + F₂+...=ma

Since the two forces are exactly opposite in direction, the answer is 40N_west + -15N_east = 25N West (the 15 is negative because west was arbitrarily chosen as the positive direction)

F_net = √((90N)²+(120N)² = 150N (the Pythagorean Theorem used to obtain the net force). Since F=ma, 150N=(5kg)a. a=30m/s².

• First split F₄ into components: F_4x= F₄cosθ=(20N)cos45°=14N and F_4y=F₄sinθ =(20N)sin45°=14N
• F_netx= 14N-10N = 4N and since F_netx=ma: 4N=(5kg)a_x, a_x=0.8m/s²
• F_nety=5N+14N-16N=3N and since F_nety=ma, 3N=(5kg)a_y, a_y=0.6m/s².

• Case 1: corresponds to speeding up
• Case 2: corresponds to slowing down
• Case 3: corresponds to a change in direction
• All other cases: correspond to both a change in direction and speeding up or slowing down.

• When one body exerts a force on another, the second body will exert an equal and opposite force on the first.
• Mathematically, F_2on1 = -F_1on2.

Newton's Third Law states that they each feel the same magnitude force. However the effect of the force is different. m₁a₁=m₂a₂. The small block will experience 100 times the acceleration of the large block.

F_g=mg Where g is the acceleration due to gravity (=10m/s)

• Weight is equal to the product:
• weight (w) = F_g = mg
• so, 50N=m(10m/s²)
• m=5kg

• First determine the object's acceleration:
• F_net=ma
• Choosing downward as the positive direction:
• 320N=80kg(a)
• a=4.0m/s²
• Knowing that the body accelerates at 4.0m/s², calculate t:
• d=v₀t+(1/2)at²
• 18m=0+(1/2)(4)(t²)
• 18m=2t²
• t²=9
• t=3s
• Knowing that the object travels for 3 seconds with an acceleration of 4.0m/s², calculate its final velocity:
• v_f=v₀+at
• v_f=0+(4.0m/s2)(3s)
• v_f=12m/s

• 
• F_g=force of gravity
• m₁= mass of object 1
• m₂= mass of object 2
• r= distance between the centers of the masses of objects 1 and 2
• G= universal gravitational constant=6.67x10¹¹N*m²/kg².
• The MCAT does not require you to know the value of the universal gravitational constant, but what is important is that gravitational force is:
• directly proportional to the mass of each object and
• inversely proportional to the square of the distance between the centers of the mass of the objects.

• Since mass is an intrinsic property of the object, it will not change. This eliminates choices B and D.
• The only variable changing in the gravitational formula is r.
• Since F_g is inversely proportional to the square of the distance between the centers of mass of the objects, it is 1/3² = 1/9 so,
• 1/9(360N)=40N. C is the answer.

The force, Fn, that refers to the force in a direction that is perpendicular to the surface.

• 1. Start by drawing all of the forces acting on the object. To find the normal force, we only need to look at Newton's Second Law in the y-direction.
• F_net,y=ma_y
• Calling upward positive, F_N-F_g=0, or m*0, since there is no acceleration in the y-direction. Therefore:
• F_N=F_g
• =mg
• =(100kg)(10m/s²)
• 2. First, split F into components. Again, we are only concerned with the y-direction.
• F_net,y=ma_y
• Calling upward positive, we see that:
• F_N+Fsinθ-F_g=m*0  Therefore,
• F_N=F_g-Fsinθ
• =mg-Fsinθ
• =(100kg)(10m/s²)-(100N)sin30°
• =1000N-50N
• =950N
• 3. Now there is an upward acceleration:
• F_net,y=ma_y
• F_N-F_g=ma_y
• F_N=F_g+ma_y
• =mg+ma_y
• =(100kg)(10m/s²)+(100kg)(2m/s²)
• =1000N+200N
• =1200N

Since the ropes, strings, or cables are massless, the two objects will experience the same magnitude of tension, but in the opposite direction.

• Examine the blocks separately. According to Newton's Second Law, the equation in the horizontal direction for block one is:
• F_T=m₁a
• For block two:
• F-F_T=m₂a
• Since the blocks move together, the acceleration is the same for each. Since they are connected by a single string, F_T is the same for each. Plugging equation 1 into equation 2, we see:
• F-m₁a=m₂a
• F=m₁a+m₂a
• F=(m₁+m₂)a
• 60N=(10kg+20kg)a
• a=2m/s²
• To solve for F_T, we can plug the value of a into either equation
• F_T=m₁a
• F_T=(10kg)(2m/s²)
• =20N

Kinetic (sliding) and static (not sliding)

• F_f,k=μ_kF_N
• where F_N is the normal force and μ_k is referred to as the coefficient of kinetic friction and is experimentally determined.

• In the horizontal direction:
• F_net=ma
• F-F_f,x=ma
• F-μ_kF_N=ma
• In the vertical direction:
• F_net=ma
• F_N-F_g=0
• F_N=F_g
• F_N=mg
• Plugging this into the horizontal eq.:
• F-μ_kmg=ma
• 200N-(0.3)(100kg)(10m/s²)=(100kg)(a)
• 200N-300N=(100kg)a
• a=-1m/s
• Note that the object is slowing down and that the friction force is greater than the applied force. A greater applied force was needed to start the object moving in the first place, but once the object was moving, the value of kinetic friction remains constant regardless of whether the applied force changes.

• F_f,s≤μ_sF_N
• where μ_s is the coefficient of static friction.

• F>F_f,s,max
• F>μ_sF_N
• F>μ_smg
• F>(0.4)(100kg)(10m/s²)
• F>400N
• Warning: Static friction does not always equal to μ_sF_N. In general, it is whatever it needs to be to make sure that the object does not move.

• The "trap" answer would be μ_sF_N, but remember this is the maximum static friction, not necessarily what static friction is in this particular case.
• F_f,s,max=μ_sF_N
• =μ_smg
• =(0.4)(100kg)(10m/s²)
• =400N
• This tells us that an applied force of more than 400N is required to start the object moving. Since the applied force is only 100N, the object is NOT MOVING.
• If the object is not moving, we can apply Newton's Second Law.
• F_net=ma
• F-F_f,s=0
• F_f,s=F=100N
• The answer is B.

• F_net=ma
• mgsinθ=ma
• a=gsinθ
• =(10m/s²)sin30°
• =5m/s²

• To make an object just start to move, the downward component of gravity must be slightly larger than the maximum static friction.
• mgsinθ>F_f,s,max
• mgsinθ>μ_sF_N
• To find the normal, we look at the y-direction:
• F_net=ma
• F_N-mgcosθ=0
• F_N=mgcosθ
• Plugging this into the above:
• mgsinθ>μ_smgcosθ
• sinθ>μ_scosθ
• tanθ>μ_s
• θ>tan^-1(μ_s)
• θ>tan^-1(0.4)
• Since we know that tan^-1(0)=0 and tan^-1(1)=45°, tan^-1(0.4) must be in between 0° and 45°. Choice A is the correct answer.

Each mass feels the same magnitude tension.

• Since each block experiences the same magnitude of acceleration, it is useful to choose the direction of motion to be positive for each block.
• Using Newton's Second Law, we see that:
• F_T=m₁a and m₂g-F_T=m₂a (if m₁ is for the black on the edge and m₂ is the block hanging off the edge)
• plugging the first equation into the second equation:
• m₂g-m₁a=m₂a (or F_net=ma, if treating the entire system as one object with mass=m₁+m₂)
• m₂g=(m₁+m₂)a
• (5kg)(10m/s²)=a(5kg+10kg)
• a=3m/s²

• In this problem, the upper pulley is fixed in place while the lower pulley moves with the mass. Treating the movable pulley and mass as an object, F_T is labeled on each side of the small pulley on the string.
• Even though it is one string lifting the pulley, the tension acts on each side of it, counting it as two forces. The minimum force required to lift the pulley-mass system is:
• 2F_T-mg=0
• F_T=1/2(mg)
• Looking at the original diagram, we can see that F is equal to F_T (since it is all on one string). Therefore:
• F=F_T
• =(1/2)mg
• =(1/2)(10kg)(10m/s²)
• =50N

 for an object moving in a circle of radius r and speed v, this is the acceleration toward the center of the circle.

• 
• 
• 
• Solving for v, we get:
• 
• Note that the mass of the satellite does not matter.

• (a) The maximum tension occurs when the ball is at the bottom of the circle, since the string needs to both support its weight and to provide necessary centripetal force. If we decide that toward the center is positive:
• F_net=ma
• F_T-mg=m(v²/r)
• F_T=mg + mv²/r
• =(5kg)(10m/s²)+(5kg+(10m/s)²)/0.5m
• =1050N
• (b) To find the minimum speed needed to make it around the circle, we look at the top of the circle.
• F_T+mg=m(v²/L)
• The minimum speed corresponds to the minimum tension (zero). Therefore:
• mg=m(v_min)²/L
• v_min=L/√(gL)
• =√((10m/s²)(0.5m))
• -2.2m/s

• For any object whose mass is uniformly distribute, the center of mass is at the geometric center.
• For a system of objects arranged along the x-axis:
• 
• A useful method to choose a location of origin is to use the location referenced in the problem. If the question asks, "how far from the left is the center of mass?" choose the left as the origin.

• (a) Conceptually, it can be seen that the center of mass will be closer to the heavier mass. This eliminates choices A and B. Choosing the left edge as the origin, we see that x=0 and x₂=L. Therefore:
• x_cm=((2kg)(0)+(6kg)(L))/(2kg+6kg)
• =3/4(L) Choice D
• (b) Even with the plank having mass, we can eliminate choices A and B. Since the mass of the plank is not given, we cannot use the formula. However, since the plank is uniformly dense, its center of mass is the actual center. This new mass would skew the answer to part (a) closer to the actual center.
• Therefore, 1/2<x_cm<3/4L, choice C is the only answer that fits the criteria.

t=rFsinθ where r is the vector that points from the pivot to the location of the applied force and θ is the angle between r and F.

• (a) Since the angle between r and F_T is 180°,
• t_T=L*F_Tsin180°θ=0
• (b) The angle between r and F_g is 30°, so
• t_g=L*F_gsin30°=L*mgsin30°= (0.5)(10kg)(10m/s²)(0.5) =25N*m

• Using F_t,s≤u_sF_N would be difficult since θ isn't given. t=lF where l is the distance from lever arm to the line of action (from the pivot point to the bottom).
• l=1/2(1m)=0.5m
• t=(0.5m)(20N)
• =10N*m

A body (or system of bodies) is in translational equilibrium if it experiences no net forces in any direction and is in rotational equilibrium if it experiences no net torque.

• Since the problem is in equilibrium, the net force on the mass is equal to zero.
• T₁-F_g=0N
• T₁=F_g=50N
• You can also examine the forces acting on the knot.
• Splitting T₃ into horizontal and vertical components, we see:
• Horizontal: T₂=T₃cos 60°
• Vertical: T₃sin60°=T₁
• Starting with the vertical equation T₃: T₁/sin60°=50N/.87=60N (approx.)
• And therefore: T2=(60N)cos60°=30N (approx.)

• While this problem can be solved using center of mass, using torque is easier. Treating the plank and blocks as one object, where mass_pl is the mass of the plank:
• Setting the clockwise torques equal to the counterclockwise torque:
• t_cw=t_ccw
• (2m)(m_pl)g+(6m)(m₂)g=(2m)(m₁)g
• (2m)(30kg)(10m/s²)+(6m)(m₂)(10m/s²)=(2m)(60kg)(10m/s²)
• 60kg*m+(6m)m₂=120kg*m
• m₂=60kg*m/6m
• =10kg

• W=Fdcosθ; where θ is the angle between F and d and is expressed in N*m, which is also called the joule (J) Torque is also expressed in N*m, but is NOT expressed in joules. This is because torque is a vector quantity. Work is scalar, and the joule is a scalar unit.
• **This is ONLY valid for constant forces!

• The mass of the object is irrelevant because the applied force is provided. Also, since the object is moving in the direction of the applied force, θ=0.
• W=Fdcosθ = (30N)(1,000m)(1)=30,000N*m
• =30,000J

• The work equals only the force applied in the direction of displacement, which is 10N times the cosine of 60°.
• Thus: W=Fdcosθ
• =(10N)(10m)(.5)=50J

• Case 1: 0≤θ<90°
•      cosθ is positive -->W is positive
• Case 2: θ=90°
•      cosθ is zero --> W is zero
• Case 3: 90°<θ≤180°
•      cosθ is negative --> W is negative

• F: since F is in the direction of d, θ=0°
•      W_F=Fdcos0°
•      =(100N)(5m)(1)
•      =500J
• F_f,k: since F_f,k is opposite the direction of d, θ=180°
•      W_f,k=F_f,kdcos180°
•      =u_kF_Ndcos180°
•      =u_kmgdcos180°
•      =(0.3)(100kg)(10m/s²)(5m)(-1)
•      =-1500J
• F_g and F_N: since these forces are perpendicular to d, θ=90°
•      =W_g=W_N=0

Kinetic friction

Static friction, all centripetal forces (since the force is directed toward the center and the direction of motion is tangent to the circle); the normal force often (but not always) does zero work (since objects often move parallel to the surface and the normal is always perpendicular to the surface)

• Since work is scalar, the total work acting on an object is merely the sum of the individual works.
• W_TOTAL=W_F+W_g+W_N+...

area under the curve

path independent

• The object is moving in the direction of F_g.
• Therefore:  W_g=F_ghcosθ
•                      =mgh
• Note that the work done by gravity can be calculated merely by looking at vertical displacement, rather than the particular path.

• The object is moving at an angle with respect to F_g. Knowing that mgsinθ is the component of gravity that points down the plane:
•      W_g=F_gd
•      =mgsinθ*L
•      =mg(Lsinθ)
•      =mgh
• Note: the work done by gravity can be calculated merely by looking at the overall vertical displacement, rather than the particular path.

In this case, we cannot use W=F_gd since the component of gravity down the plane changes. However, since gravity is a conservative force, we can say that W_g=mgh.

• Since there was no vertical displacement,
• W_g=0

KE=(1/2)mv² with the units in joules (J)

KE=(1/2)mv²=(0.5)(40kg)(20m/s)²=8,000J

• W_TOTAL=ΔKE=KE_f-KE₀
• The total work done on an object equals the object's change in kinetic energy.
• Conceptually, this means that if a force does positive work on an object, the object gains KE. If a force does negative work on an object, the object loses KE.

• We know that W=Fdcosθ, but no information was given about F, d, or θ. We do not need to know what is actually stopping the car (friction? a wall?), but we can calculate the work done based on how much kinetic energy is being lost.
• W_TOTAL=KE_f-KE_o
• =(1/2)(1,000kg)(0)²-(1/2)(1,000kg)(20m/s)²
• =-200,000J
• We expect the answer to be negative since kinetic energy is being taken away from the car.

• We cannot solve this with kinematics since the Big Five require uniform acceleration. Instead, we can use the Work-Kinetic Energy Theorem:
• W_TOTAL=KE_f-KE_o
• Since the ramp is frictionless, the only forces acting on the object are gravity and the normal. The work done by the normal force is zero. Therefore:
• W_g=KE_f-KE_o
• =>mgh = (1/2)mv_f²-(1/2)mv_o²
• =>mgh=(1/2)mv_f²-0
• =>v_f=√(2gh)
• This problem can also be approached using Conservation of Mechanical Energy:
• KE_o+PE_o=KE_f+PE_f
• 0+mgh=(1/2)(mv_f²)+0
• solving for v_f=√(2gh)

• PE=mgh
• A classic example of potential energy is a stretched or compressed spring; the spring has the potential to perform work.
• For an object falling from height h, we can say that the potential energy of the system was converted into kinetic energy.
• Note: the ground is an arbitrary point of reference - h can be measured from anywhere

• (A) PE=mgh, where h is 30m:
• =(75kg)(10m/s²)(30m)
• =22,500J
• (b) PE=mgh, where h is now 0:
• =(75kg)(10m/s²)(0)
• =0
• (c) PE=mgh, where h is now -30m:
• =(75kg)(10m/s²)(-30m)
• =-22,500J
• Note: the actual value of potential energy isn't important. How it changes it the important part. We can say that W_g=-ΔPE

• constant
• This means that the energy in the universe may change form, but it cannot be destroyed or created. MCAT questions will sometimes draw on the candidate's ability to apply this principle to situations and systems involving the conversion of energy from one form to another.

• E=KE+PE
• We know that W_TOTAL=ΔKE
• If gravity is the only force doing the work, then W_g=ΔKE.
• We also know that W_g=-ΔPE
• Therefore: -ΔPE=ΔKE
• or: Δ(KE+PE)=0
• or: KE_o+E_o=KE_f+PE_f

• While the problem could be solved using kinematics, it would be very time-consuming. Conservation of Mechanical Energy is much easier:
• KE_o+PE_o=KE_f+PE_f
• Setting the ground as the reference point:
• (1/2)m(10m/s)²+m(10m/s²)(40m)=(1/2)mv_f²+0
• canceling m from each term:
• 50m²/s²+400m²/s²=(1/2)v_f²
• =>v_f=√[2(50m²/s²+400m²/s²)]
• =30m/s

Conservation of Energy; if F is the non-conservative force, you may use KE_o+PE_o+W_F=KE_f+PE_f or W_F=Δ(KE+PE)

• The word "required" means that we are looking for the minimum work and, hence, the minimum force. The long way would be to solve for F using W-Fdcosθ. A shorter way is to look at work and energy. Since the force will not be accelerating the block, the kinetic energy will remain constant. Therefore, the work done by F is only giving the block potential energy.
• PE_o+W_F=PE_f
• Setting the ground as the reference:
• W_F=mgh
• =(10kg)(10m/s²)(5m)
• =500J
• Note: this is the same as if the force were applied directly up the plane or if the force were lifting the object straight upward.

• To find the length of the inclined plane:
• sin30°=12m/d
• d(sin30°)=12m
• d(0.5)=12m
• d=24m
• If an 80N object is to be lifted a distance of 12 meters, the work required is equal to:
• Fd=W
• (80N)(12m)=960J
• To find the F needed to travel the inclined plane of 24m (as found previously) then:
• W=Fd
• 960J=F(24m)
• F=40N
• To lift the object straight up would take 80N and to have the mechanical advantage of the inclined plane would take a force of 40N

• P=W/Δt (work over the time). It is expressed in J/s or watts (W)
• P=Fvcosθ since watts is equivalent to N*m/s which is the product: (force)(velocity)

• Since the object is not accelerating, the upward force necessary to lift it is merely equal to the object's weight. The object's mass is 75kg, and its weight is:
• F_g=weight=75kg(10m/s²)=750N
• The lift will involve the work of:
• W=Fd=(750N)(16m)=12,000J
• If 12,000 joules of work are to be performed in 20 seconds, the associated power would be:
• P=W/t=12,000J/20s=600W
• Note: watt is equivalent to N*m/s, which also represents the product: (force)(velocity).

• Since the force is acting in the same direction of the motion:
• P=Fvcosθ where cosθ=1
• 27,000W=F(300m/s)
• F=90N
• The force applied by the engine to keep the object moving upward at constant velocity=90N, which if we assume no air resistance, means the object's weight =90N. This means its mass is:
• F=ma where a = gravity
• 90N=m(10m/s²)
• m=9kg

• p=mv  (mass of body)(velocity) 
• and is expressed in units of (kg)(m/s) or kg*m/s

• p=mv
• (100kg)(20m/s)=2,000kg*m/s

• Impulse is denoted by J
• F_av*Δt=Δp  or J=Δp
• Impulse can be measured in N*s or kg*m/s

• (A) Momentum, like velocity, is a vector. If the direction of the initial velocity of the ball is considered positive:
• p_o=mv_o=(0.15kg)(40m/s)
•      =+6kg*m/s
•  and p_f=mv_f=(0.15kg)(-40m/s)
•      =-6kg*m/s
• Therefore, the change in momentum is
• Δp=p_f-p_o=(-6kg*m/s)-(6kg*m/s)
•      =-12kg*m/s
• Note: since direction matters, the answer is not zero.
• (b) J=Δp=-12kg*m/s or -12N*s
• (c) Since, by definition,
• J=F_avΔt
• -12kg*m/s=F_av(15x10^-3s)
• F_av=-800N
• The magnitude of the force is therefore 800N (the answer was positive because the bat is hitting the ball in the negative direction)

• m₁v_1o+m₂v_2o=m₁v_1f+m₂v_2f
• This is known as the Law of Conservation of Momentum.

• Choosing the positive direction to the right:
• m₁v_1o+m₂v_2o=m₁v_1f+m₂v_2f
• (2kg)(6m/s)+(1kg)(0)=(2kg)(3m/s)+(1kg)v_2f
• v_2f=6 m/s

• For all collisions between objects free of net external forces, momentum is conserved.
• For collisions where kinetic energy is conserved, then
• Elastic collision: p_o=p_f and KE_o=KE_f
• If kinetic energy is NOT conserved, then
• Inelastic collision: p_o=p_f and KE_o≠KE_f

• Choosing the positive direction to the right:
• m₁v_1o+m₂v_2o=m₁v_1f+m₂v_2f
• (2kg)(6m/s)+(1kg)(0)=(2kg)(3m/s)+(1kg)v_2f
• v_2f=6 m/s
• KE_o=(1/2)m₁v_1o²+(1/2)m₂v_2o²
• =(1/2)(2kg)(6m/s)²+(1/2)(1kg)(0)²
• =36J
• KE_f=(1/2)m₁v_1f²+(1/2)m₂v_2f²
• KE_f=(1/2)(2kg)(3m/s)²+(1/2)(1kg)(6m/s)²
• =9J+18J
• =27J
• Since KE_o≠KE_f the collision is inelastic.

• Since the objects move together after the collision, we can write:
• m₁v_1o+m₂v_2o=(m₁+v₁)v_f
• Choosing the positive direction to the right:
• (2kg)(6m/s)+(1kg)(-3m/s)=(2kg+1kg)v_f
• v_f=+3m/s (to the right)

First, a picture:

• 
• Since the surface is frictionless, use Conservation of Momentum. Choosing the positive direction to the right:
• m_pv_po+m_bv_bo=m_pv_pf+m_bv_bf
• 0+0=(60kg)v_pf+2kg)(-10m/s)
• v_pf=+0.3m/s
• To determine if it is elastic, look at initial and final kinetic energies:
• KE_o=(1/2)m_pv_po2+(1/2)m_bv_bo²
• =0+0=0
• KE_f=(1/2)m_pv_pf²+(1/2)m_bv_bf²
• =(1/2)(60kg)(0.3m/s)²+(1/2)(2kg)(-10m/s)²
• =2.7J + 100J
• =102.7J
• Since KE_o≠KE_f, the situation is inelastic.
• Note: that KE_f>KE_o means that the system gained kinetic energy, as opposed to a perfectly inelastic collision, where kinetic energy was lost.

• a measure of mass per volume and is expressed in the SI unit kilogram per cubic meter (kg/m³).
• Density=mass/volume

• density = mass/volume
• 9.3x10⁴=88kg/volume
• volume = 9.5x10^-4m³

• The density of water is 1,000 kilograms per cubic meter.
• Specific gravity represents the ratio of density of the solid (or substance) to density of water.
• 
• So, this solid has a specific gravity of 2.

• The density of water is 1,000 kilograms per cubic meter.
• Specific gravity represents the ratio of density of the solid (or substance) to density of water.
• 600/1000=0.60

• Density = mass/volume
• The object's mass is given as 0.2 kg.
• The object's volume is equal to the volume of the liquid it displaces:
• (0.75x10^-3m³)-(0.50x10^-3m³)=0.25x10^-3m³
• Density: 0.2kg/0.25x10^-3m³=
• (0.2x10⁰kg)/(0.25x10^-3m³)=
• 0.8x10³kg/m³
• Specific gravity: 800/1000=0.8

• A force that (a) is applied equally at both ends of a solid object and (b) tends to stretch or compress it.
• Equals force/cross-sectional area
• Measured in pascals (Pa) where 1N/m²=1Pa

• Rod's cross-sectional area: =(3.14)(0.03)²=2.83x10^-3m² 
• Tensile stress experienced:
• force/cross-sectional area=
• 90N/2.83x10^-3m²=
• 90x10⁰N/2.83x10^-3m²=
• 31.80x10³Pa

• T_ss=F/A_cs 
• Cross-sectional area:
• (0.04m)(0.03m)=0.0012m²=1.2x10^-3m² 
• Stress:
• 40N/1.2x10^-3m²=
• 40x10⁰N/1.2x10^-3m²=
• 33.33x10³Pa

• The degree to which tensile stress causes a solid object to undergo a change in length.
• (change in length)/(original length) = ΔL/L_o
• No units are used as strain is a dimensionless quantity.

• ΔL/L_o
• =(2.01m-2.0m)/2.0m
• =0.01m/2.0m=.005

• ΔL/L_o
• =(34m-33.97m)/34m
• =0.03m/34m=0.00088 or 8.8x10^-4

• Every substance that obeys Hooke's law is associated with its own Young's modulus--a value that, for that particular substance, expresses the degree of strain that will be imposed by a given degree of stress.
• For any object that obeys Hooke's law (force, F, needed to extend or compress a spring by some distance, d, is proportional to that distance. That is, F=kd), it is the ratio:
• tensile stress/tensile strain
• It is expressed in pascals (Pa)

• 50 Pa/tensile strain=7.5x10⁵ Pa
• tensile strain = 50 Pa/7.5x10⁵ Pa
• =6.67x10^-5
• The problem does not specify whether the imposed stress was stretching or compressive; the math is the same in either case.

• Calculate cross-sectional area:
• =(3.14)(0.015)²=0.00071m²=7.1x10^-4m²
• Calculate strain:
• ΔL/L_o=(5m-4.9981m)/5m=0.0019m/5m
• =0.00038=3.8x10^-4
• Calculate stress:
• stress/strain=5.2x10⁶Pa
• stress=(5.2x10⁶Pa)(3.8x10^-4)
• =1.9x10³Pa
• Calculate force:
• T_ss=F/A_cs
• F=(T_ss)(A_cs)=(1.9x10³Pa)(7.1x10^-4m²)
• F=1.4N

• Pressure=(density)(acceleration due to gravity)(depth below the surface of the fluid)
• P=ρ(g)(h)
•  where ρ=fluid density,
•  g=acceleration due to gravity
•  h=height of the fluid above the point in question

• Gauge pressure: P=ρ(g)(h)
• Absolute pressure: P=ρ(g)(h)+atmospheric pressure. (or gauge pressure plus atm pressure)
• Absolute pressure is the actual pressure to which a submerged object is exposed.

• Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
• It follows that fluid pressure at any given depth in a resting fluid is unrelated to the shape of the container in which the fluid is situated.

• Because fluid pressure is directly proportional to fluid density and depth, the pressure at depth 2h in fluid E is calculated:
• P=(3000Pa)(0.4ρ)(2h)
• =0.8(3,000Pa)
• =2,400 Pa

• Buoyancy (F_b)=(V)(ρ)(g)
• =(volume of fluid displaced)(density of displaced fluid)(acceleration due to gravity)
• Buoyancy (F_b)= weight of the fluid displaced

• (1) The only forces to which the object is subject are those of its own weight, and the upward for of buoyancy.
•      Object's weight:
•       w=57kgx10m/s²=570N
•      For buoyancy calculate volume:
•       volume = mass/density
•       volume = 57kg/11.3x10³kg/m³
•           =5.0x10^-3m³
• Knowing that the solid displaces 5.0x10^-3 cubic meters of water and using the known density of water (1,000 kilograms per cubic meter)...
•      Buoyancy=(V)(ρ)(g):
•       =(5.0x10^-3m³)(1x10³kg/m³)(10m/s²)
•       = 50N
• So, the object experiences a downward force of 570 newtons and an upward force of 50 newtons. It experiences, therefore, a net force of 520 newtons downward because:
•      F_net=50N-570N=-520N
• (2)The object will accelerate downward. Since its downward net force has a magnitude of 520N:
•       F=ma
•      520N=57kg(a)
•      a=approx. 9.1m/s²

• Flow=(area of pipe or tube)(velocity of fluid)
• Ideal flow:
•    is constant at all points,
•    has no viscosity,
•    has no turbulence,
•    has no drag (fluid friction), and
•    =(area)(velocity)

• Laminar flow refers to flow in which whirlpools and eddies are absent.
• Turbulent flow is flow in which whirlpools and eddies are present.

Velocity is inversely proportional to cross-sectional area

An increase in one corresponds to an increase in the other, but the relationship is not proportional.

• (1) Where caliber is greatest, pressure is greatest; thus pressure is greatest at section A.
• (2) Where caliber is lease, velocity is greatest; thus velocity is greatest at section C.
• (3)(a) Area and velocity are inversely proportional. Ascertain the ratio. area of section B : area of section D
• Looking to the formula for the area of a circle, the area of the circle is proportional to the square of the radius. The radius of section B is 3 meters, and the radius of section D is 1 meter. The two radii bear the ratio 3:1, which means that the ratio of their areas is:
•      3²:1² = 9:1
• Since velocity is inversely proportional to area, and the velocity in section B is 0.5 m/s, the velocity in section D is:
•      (0.5m/s)(9)=4.5m/s
• (b) Since flow equals the product of area and velocity; flow of section D equals
•      x(4.5m/s)=(3.14)(1m)²x(4.5m/s)
•       =14.13m³/s
• (c) Flow is equal at all points. Having calculated flow in section D, section A will have the same flow of 14.13m³/s

• A poor conductor has relatively poor electron mobility.
• A poor conductor will neither produce nor experience contact charge.

A is correct. Contact charge between a charged and uncharged body involves the movement of electrons (NOT PROTONS) from one body to another and proceeds only if both bodies experience relatively good electron mobility; in other words, both bodies must be good conductors.

D is correct. The question concerns contact charge as applied to a negatively and positively charged body. Both bodies are good conductors. Body X will impart electrons to body Y, which makes choice III correct. The negative charge on body X will be reduced in magnitude, as will the positive charge in body Y. Choices I and II are also accurate. Note that simply knowing the direction of electron flow allows you to choose III, and by extension D.

A is correct. The question concerns charge by induction. Object A neither gains nor loses electrons throughout the process and so maintains its original negative charge (it was brought near but was no touching object B). Some of the electrons in object B are repelled by the negative charge of object A and passed to object C. When object C is removed, object B is left positively charged

• The problem draws on the proportionalities associated with Coulomb's law.
• Coulomb's law:
•  where q1=charge on object 1
•            q2=charge on object 2
•            r=distance between the centers of the charge
•            the first fraction is a constant
• The repulsive (or attractive) force created by two point charges is inversely proportional to the square of the distance between them.
• Distance is reduced by a factor of:
• 3/0.75=4
• repulsive force increased:
• 4²=16
• So the equation that expresses F_r2:
• F_r2=16(F_r1)

• As noted, A and B experience an attractive force, since their charges are of opposite sign, and particles G and H experience a repulsive force, since their charges are of like sign.
• The question concerns the relationship between the magnitude of these forces, and the answer lies in the proportionalities associated with Coulomb's law:
• F=k[(q₁q₂)/r²]
• In relation to the first pair of particles, the second pair of particles generates a numerator that is:
• (3)x(6)=18 times as large.
• However, the second pair of particles is separated by one half the distance that separates the first pair. Since the applicable force is inversely proportional to the square of the distance between the particles:
• F_a(ab)=[1/2²](F_r(gh))=[1/4](F_r(gh))
• Combine the effects of the numerator and denominator:
• F_a(ab)=(1/4)(1/18)(F_r(gh))
• Fa(ab)=F_r(gh)/72

• According to Coulomb's law (F=k[q₁q₂/r²]), tripling the magnitude of charge on one of the  particles will triple the repulsive force between the particles. Tripling the distance between the particles reduces the force by a factor of:
• (3)²=9.
• Multiply the new force by 3/9 or 1/3:
• 33N*(1/3)=33/3=11N

• The problem requires the use of the mathematical relationships inherent in Coulomb's law:
• F=[1/(4πε₀)][q₁q₂/r²]
• (a) The force between two charged particles is inversely proportional to the square of the distance between them. The ratio 6y:y is equivalent to the ratio 6:1. Since (6)²=36, the electrostatic field strength at P₂=x/36 = 0.008/36 N/C = 2.2x10^-4 N/C.
• (b) C is correct. Using Coulomb's law:
• 0.008N=(1/4πε₀)[(x*1)/y²]
• Solving for x:
• 0.008÷[1/4πε₀]=(x*1)/y²
• (0.008)(4πε₀)=(x)/y²
• x=y2(0.008)(4πε₀)

• Force_es equals (electrostatic field strength)(charge on particle) or:
• F_es=E_fs(q)

• F_es=(electrostatic field strength)(charge on particle)
• F_es=E_fs(q)
• F_es=(6 N/C)(0.0022 C)
• F_es=0.0132 N

• Charge on one plate
•  (ε₀) (area of plate)
• or
• E=Q/[(ε₀)(A)]

• The charge on each plate in the first pair in relation to the charge on each plate in the second pair gives rise to the ratio 1:1.2 = 5/6.
• The area of each plate in the first pair in relation to the area of each plate in the second pair gives rise to the ratio 3:3.6 = 5/6.
• This second ratio is the ratio of the areas, which is inversely proportional to the electric field and, thus, should be inverted. The electric field strength associated with the first pair of plates in relation to that associated with the second creates the ration (5/6):(5/6)=1.
• The equation for the electrostatic field, E₁, in terms of E₂ is simply E₁=E₂.

• Electrostatic potential = (Electrostatic field strength) (distance between plates)
• V=Ed
• calculated in volts
• 1V=1J/C

800Nx(.015m)=12N*m=12J

• Recall that:
• 1. electrostatic field strength (E) is directly proportional to the charge on either plate (Q) and inversely proportional to the area of either plate (A), and
• 2. electrical potential (V) is directly proportional to electrostatic field strength (E) and also directly proportional to the distance between the plates (d).
• V is directly proportional to the fraction:
• (Q)(d)
•     A
• The doubling of (A) tends to reduce V by a factor of 2, and the tripling of d tends to increase V by a factor of 3. The alterations, therefore, multiply V by the fraction 3/2 = 1.5, which means that:
• V₂=(1.5)(V₁)

• The term capacitance refers to the degree to which a capacitor stores an electric charge in relation to its electrostatic potential. For any capacitor, capacitance (C) is proportional to the quotient:
• (charge)/(electrostatic potential) or
• Q/V
• expressed in C/v (Coulombs/volt) or a farad

• 1. the area of the plates is inversely proportional to electric field strength (E).
• 2. The electric field strength is directly proportional to electrostatic potential (V).
• 3. The electrostatic potential is inversely proportional to capacitance (C); increased size of the capacitor's plates is directly proportional to capacitance.
• In terms of algebra:
• C is proportional to Q/V
• V is proportional to (E)(d)
• E is proportional to Q/(ε₀)(A)
• C is plainly proportional to:
• Q÷ (d)(Q) = (Q)(ε₀)(A) = (ε₀)(A)
•      (ε₀)(A)       (d)(Q)          d
• which is proportional to A
• So, if you have the dielectric constant...
• C= [(K)(ε₀)(A)]/d

• A. is correct. The designation "100-volt capacitor" signifies a capacitor in which a hypothetical particle with charge of magnitude 1 coulomb would acquire 100 joules of potential energy when moved from the plate with a charge of opposite sign to the plate with a charge of like sign.
• 100V= 100J/C
• 100J/Cx3x10^-7C=300x10^-7J=3.0x10^-5J

• Knowing that the electrostatic field strength is equal to the fraction:
• (Q)/[(ε₀)(A)]
• Calculate:
• 9x10^-4C/[8.85x10^-12C²/N*m²)(9x10^-3m²)=
• 90x10^-5C/79.65x10^-15C²/N=
• 1.13x10¹⁰N/C

D. is correct. If the movement of a charged body between two points in an electrostatic field requires work, then the body is acquiring potential energy equal to the amount of work required minus any energy that's converted to other forms, such as heat. In no event can the body acquire potential energy greater than the work performed on it. When the charged body is allowed to move back to its original position, its potential energy might be harnessed to perform work. The maximum amount of work that might be performed is equal to the potential energy, which, in turn, is equal, at maximum, to the amount of work originally performed to move the object from its original position. If the work done by the object during its "return" trip is less than that done on the object during its initial movement, the explanation might be that the potential energy it acquired has been converted, in part, to heat, and not directed to the production of work.

• A. is correct. The investigator intends to apply the energy stored within a capacitor (the potential energy represented by its electrical potential) toward the performance of work. The lifting of a 5 kilogram body 2 meters upward imparts gravitational potential energy of:
• (m)(g)(h)=(5kg)(10m/s²)(2m)=100 joules
• which corresponds to 100 joules of potential energy. Since the operation of the machine itself will require the performance of work and hence the conversion of potential energy, it is plausible that the actual amount of potential energy lost by the capacitor will be greater than that acquired by the lifted object.

• I=V/R
•   where
• I= current (amperes-A)
• V= voltage (volts-v), and
• R= resistance (ohms-Ω)
• Current is directly proportional to voltage (electrical potential difference) and inversely proportional to resistance.

• I=V/R
• 12v/2Ω
• =6A

• I=V/R
• 4800A=1800v/R
• R= 1800v/4800A
• =0.375Ω

• I=V/R
• 40A=50v/R
• R=1.25Ω

• The problem requires an appreciation of the proportionalities of Ohm's law. Current is directly proportional to voltage and inversely proportional to resistance. Since resistance is multiplied by a factor of 3, and voltage by a factor of 1/2, the new current, I₁, comes from the original current, I, divided by a factor of 3 and multiplied by a factor of 1/2:
• I₁= (I)x[(1/2)/3]
• I₁= I/6

• 1. P=IV
• 2. since V=IR; P=I(IR) or P=I²R
• 3. since I=V/R; P=(V/R)(V) or P=V²/R

• P=IV
• 200W=(1.60A)(V)
• V=125-volts

•      work   force x dist   (mass x accl)x dist
• P= time =     time      =       time
• and
• P=IV

• P=work/t
• P=V²/R
• Power=(3000V)²/1200Ω
•   =7500W
• which by definition means
•   =7500 J/s
• Knowing W=(force)(distance)
• =[(200,000kg)(10m/s²)][(8m]
•   =1.6x10⁷J
• Then solve for time...
• 7500J/s=(1.6x10⁷J)/time
• time=2133s
•   approx. = 36 minutes

• Simply sum the resistances of each of the resistors:
• R_t=R₁+R₂+R₃+...

• Total resistance is equal to the reciprocal of the sum of reciprocals of individual resistances.
• 1/R_t=(1/R₁)+(1/R₂)+(1/R₃)+...

• (a) The total resistance within the circuit is the sum of the four resistors connected in series, including, as one resistor, the equipment itself, and as another resistor, the set of resistors connected in parallel. The resistors other than those connected in parallel have a total resistance of:
• 3Ω+2Ω+3Ω=8Ω
• The set connected in parallel features three resisting entities, each composed of two resistors in series.
• The series on the right has a resistance of 6Ω+18Ω=24Ω;
• the middle series, 5Ω+3Ω=8Ω
• and on the left, 5Ω+7Ω=12Ω
• The entire set, therefore, is equivalent to three parallel resistors of 24Ω, 8Ω, and 12 Ω, respectively.
• To determine the total resistance of the parallel resistors:
• 1/R_t=1/24Ω+1/8Ω+1/12Ω
•   =1/24Ω+3/24Ω+2/24Ω
•   =6/24Ω  = 1/4Ω
• The reciprocal of 1/4 is 4, so the total resistance of the set is equal to 4Ω.
• The total resistance of the circuit is 4Ω+8Ω=12Ω
• (b) Knowing now that the circuit's total resistance is 12Ω, the current may be calculated using Ohm's law:
• V=IR;  220V=I(12Ω)
• I=18.33A
• ((c) knowing that the circuit's resistance is 112Ω, and its current is 18.33 amperes, power can be calculated using  P=I² or P=IV
• P=IV; P=(18.33A)(220v)  =4,032 watts

• (a) The circuit features two sets of parallel resistors. The total resistance of the circuit is equal to the sum of all resistors connected in series, which means the sum of (1) the first set of parallel resistors, (2) the equipment with resistance of 7 ohms, (3) the 3-ohm resistor, and (4) the second set of parallel resistors. The total resistance of the first set of parallel resistors is:
• 1/R_t=1/40Ω+1/40Ω
• 1/R_t=2/40Ω
•   R_t=20Ω
• Total resistance of the second set of parallel resistors is determined first by summing two resistors connected in series:
• R=22Ω+23Ω = 45Ω
• The parallel set is thus equivalent to a set consisting of a 45-ohm resistor, a 90-ohm resistor, and a 30-ohm resistor. Total resistance of the set is:
• 1/R_t=1/45Ω+1/90Ω+1/30Ω
•   =2/90Ω+1/90Ω+3/90Ω
•   =6/90Ω
•   R_t=90Ω/6  =15Ω
• Total resistance of the circuit is equal to
•   20Ω+7Ω+3Ω+15Ω = 45Ω
• (b) Knowing that the circuit's total resistance is 45Ω, its current is calculated by reference to Ohm's law:
• V=IR => 450V=I(45Ω)
•  I=10A
• (c) Knowing that the circuit's current is 10 amperes, the associated power is calculated according to the equation, P=I²R, or P=IV
• P=IV => P=(10A)(450V) = 4500W

• The total resistance = 20Ω+7Ω+3Ω+15Ω=45Ω
• The current: V=IR; 450V=I(45Ω); I=10A
• The total resistance between point A and B is 20Ω.
• The voltage drop: V=IR =(10A)(20Ω)=200V

• Parallel resistors total: 1/R_t=1/24Ω+1/8Ω+1/12Ω
•    =1/24Ω+3/24Ω+2/24Ω
•    =6/24Ω =1/4Ω
• R_t=4Ω
• Current: V=IR; 220V=I(12Ω); I=18.33A
• A and B: V=(18.33A)(3Ω) = 55V
• A and C: V=(18.33A)(5Ω) = approx. 92V
• B and C: V=(18.33A)(2Ω) = approx. 37V
• C and D: V=(18.33A)(4Ω) = approx. 73V
• A and D: 55V+37V+73 = 165V
•      or V=(18.33A)(9Ω) = 165V
• Note: if the 165-voltage drop between A and D is added to the voltage drop between the negative terminal itself, and point D, where a 3Ω resistor is interposed, the result is
• 165V + [(18.33A)(3Ω)=165V+55V=220V

T= N/A*m

• The right-hand rule:
• Imagine a current-a stream of positive charge-flowing through a conducting wire...
• ___________________________
• (-)  <-------------- I(current)  (+)  
• position the right hand so that the thumb points in the direction of the current in the wire. If the hand grasps the wire in this way, the bend of the fingers around the wire will indicate the direction of the induced field surrounding the wire.

C. is correct. Using the right-hand rule, the fingers curl upward, then toward the reader over the top of the wire, then downward to curl underneath, with the thumb pointing toward the right

• F=qvB
• Force=(charge of particle)(velocity)(magnetic field strength)
• Another right-hand rule: with the thumb pointed in the direction of the current, the direction the palm is facing is the direction of the force.

• Since you know the direction of deflection, you can place your right hand with your palm facing the direction indicated for the force. Next point your thumb in the direction of the current (opposite electron flow). Your fingers should be pointing to the left. This is the direction of the magnetic field.
•         /
• <-----|------
• <-----|------
•          * e^- path

C. is correct. Strength is inversely proportional to r, the distance from the conducting wire. If at distance r, field strength equals B, then at a distance of 2.5r, the field strength will equal 1/2.5(B) or (0.4)B.

• Period: time/cycle seconds (a wave that has a period of 4 seconds is one that undergoes one cycle in 4 seconds)
• Frequency: cycle/second (if a wave undergoes 1 cycle in 4 seconds, then it  undergoes 0.25 cycles in one second. It is expressed in hertz (Hz) which is the same as "cycles per second".

• (a) The appearance of 5 crests in a period of 8 seconds indicates that the wave has undergone 4 (not 5) cycles in 8 seconds. The first crest does not represent a cycle. Rather, the appearance of 2 crests represents a cycle, and the appearance of 5 crests represents 4 cycles.
• The wave undergoes 4 cycles in 8 seconds. The fraction (8 seconds)/(4 cycles) is equivalent to the faction (2 seconds)/1, and so
• Period: the period of the wave is 2 seconds.
• (b) The wave's period is 2 seconds, and so it undergoes 1 cycle in 2 seconds. The frequency is how many cycles it undergoes in 1 second, which is the same as the reciprocal of the period.
• Frequency: 1 cycle/2 seconds=0.5 cycles/second = 0.5 Hz.

• The second of the 13 troughs that pass the observer's view represents the completion of 1 cycle, and the 13 troughs in total represent the completion of 12 cycles. The wave, therefore, undergoes 12 cycles in 60 seconds.
• (a) The wave's period is derived from the fraction 60 seconds/12 cycles = 5 seconds/cycle, which is conventionally expressed as 5 seconds.
• (b) The wave's frequency is the reciprocal of its period. The reciprocal of 5/1 is 1/5=0.2, and the frequency, therefore, is 0.2 Hz.

• Wavelength: the distance from one crest to the next (or from one trough to the next) and is symbolized by the Greek letter lambda (λ).
• Speed: (frequency)(wavelength) = fλ m/s

• a) If period = 4x10^-5s, frequency = 1/(4x10^-5)=0.25x10⁵ or 2.5x10⁴Hz
• which means it undergoes 25,000 cycles per second
• b) v=fλ; 680m/s=λ(25,000Hz)
•    =2.72x10^-2m

Amplitude represents the wave height; the distance from the central portion of one wave unit to the crest or the distance between the central portion of the wave unit and the trough.

a wave that travels parallel to the plane in which its own medium oscillates and gives rise cyclically to areas of increased and decreased density

Since the overall mass of the bar through which a compression will travel longitudinally does not change, each region of increased density must leave behind an area of reduced density. The area of reduced density is called an area of rarefaction, and for a longitudinal compression wave, the analogs to crest and trough are compression and rarefaction, respectively.

• a) Knowing the wave has a period of 0.0286 seconds (meaning it undergoes one cycle in 0.0286 seconds), identify its frequency by taking the reciprocal of that number:
• 1/0.0286 = approx. 3.5x10¹Hz or 35Hz
• b) Knowing that the wave's frequency is 35 Hz (meaning that it undergoes 35 cycles in one second) and that its speed is 1,190 meters per second, its wavelength (which is in reality meters/cycle) is calculated by the equation:
• speed=(frequency)(wavelength)
• 1190m/s=(35Hz)λ
• λ= 34m

B. is correct. Since the two media are equally resistant to compression, density is the only factor affecting speed. The higher the density of the medium, the slower the conduction.

C. is correct. The speed of sound is dependent not only on the medium's resistance to compression but also on its density.

• Loudness (β) = decibel (dB)
• Intensity (I) = watts per square meter (W/m²)
• β=10 log (I/I₀)
•    where β is loudness in decibels, I is intensity in watts per square meter, and I₀ is the threshold of human hearing = 10^-12 watts per square meter.
• NOTE: If any given intensity, I₁, corresponds to a particular loudness, β₁, then for any x>1, a new intensity I₂, which is equal to (I₁)(10^x), corresponds to a new loudness, B₂, which is equal to (β₁+10x).
• So,
• Intensity (Z W/m²)                Loudness
• 100(Z) W/m²  corresponds to  Y+20dB (x=2)
• 1,000(Z)W/m² corresponds to Y+30dB (x=3)
• 10000(Z)W/m² corresponds to Y+40dB(x=4)

C. is correct. For any x>1, a new intensity I₂, which is equal to (I₂)(10^x) corresponds to a new loudness, β₂, which is equal to β₁ + 10x. In this instance, loudness has be increased by 20, which is equal to 10(2). The x value, therefore, is 2. Intensity, therefore, is multiplied by 10^x, which in this instance, is 10²=100.

A. is correct. The x value as used in this context is 3. When intensity is multiplied by 1,000, loudness is increased by 10x, or 30.

• 
• f_o = frequency perceived by the observer,
• f_s = frequency as actually emitted,
• v = the speed of the wave in the given medium (usually air),
• v_o = velocity at which the observer moves, which is (+) if the observer moves toward the source and (-) if the observer moves away from it, and
• v_s = velocity at which the source moves, which is (-) if the source moves toward the observer and (+) if it moves away from the observer.

• The observer is stationary, and the sound source is moving toward the observer. The numerator of the fraction is v=340m/s, and the denominator is (340m/s-80m/s). The subtraction of 80 from 340 increases the size of the fraction, wince the sound source is approaching the observer. The equation becomes:
• f_o=(400Hz)x(340m/s)/(340m/s - 80m/s)
• f_o=(400Hz)(340m/s/260m/s)
•   =400Hz(1.3)= approx. 520 Hz

• Both sound source and observer are in motion. The sound source moves away from the observer, which means the sign within the denominator will be (+), reducing the size of the fraction. The observer moves toward the sound source, which means the sign in the numerator will also be positive (+), increasing the size of the fraction:
• f_o=(280Hz)x[(340m/s+70m/s)/(340m/s +30m/s)]
•    =(280Hz)(410m/s/370m/s)
•    =280Hz(1.1)= approx. 310 Hz

• Hooke's law:
• F=-k(x)
• where
•    F=force
•    k=spring constant (expressed N/m), and
•    x=displacement
• Note: In the case of a pendulum, the spring constant is approximately equal to the fraction (weight of pendulum)/(length of pendulum)

• The problem calls for the application of Hooke's law
• F=-k(x)
• k is given as 2.5x10³ N/m
• x is given as 0.06 meters
• F=(-2500 N/m)(0.06m)= -150N
• The (-) sign in the Hooke's law formula indicates that force is opposite the direction of displacement. The force at issue is restorative; it tends to restore the moving object to the point at which it has no potential energy. The force of a swinging pendulum is downward as the pendulum moves upward. The force of a spring is toward the relaxation point as the spring moves away from it.

• Upward end:
• PE=mgh=(25kg)(10m/s)(2m)=500 joules (J)
• Bottommost position:
• When the pendulum reaches the bottommost position of its path, all of its potential energy has been converted to kinetic energy, potential energy is zero, and kinetic energy is at a maximum
• NOTE: Since the total sum (kinetic energy)+(potential energy) is unchanged at all points in the path, the sum of (mgh)+1/2(mv²) is constant throughout the path of simple harmonic motion.

• In the case of a swinging pendulum, the potential energy to be considered is gravitational potential energy = mgh. Kinetic energy = 1/2mv². The sum: (mgh)+(1/2)(mv²) is constant throughout the path of simple harmonic motion. At the bottommost position, potential energy is zero, and the entire sum of potential and kinetic energy is attributable to kinetic energy alone:
• 1/2(mv²)=1/2 x (240kg) x (8²) = 7,680J
• At the uppermost portion of the pendulum's path, kinetic energy is zero and potential energy, therefore, must be equal to 7,680J. Knowing that (mgh)=7,680 joules, you can then solve for h,
• mgh=7680J; (249kg)(10m/s²)h=7,680J
•   2,400h=7,680J
•   h=3.2 meters

• D. is correct. Each sound source moves toward the observer at a rate of 90 m/s. The perceived frequencies will be greater than the true frequencies emitted by the instruments. The factor by which each frequency is multiplied to yield the apparent frequency is equal to the fractional component of the Doppler formula:
• f₀=f_s [(v±v₀)/(v±v_s)
• Since the observer does not move, the sum (v±v_o)=340m/s. Since the sound source is moving toward the observer, frequency will increase, and the denominator must be reduced to reflect an increase in the overall size of the fraction. The sign employed in the denominator, therefore, must be (-), and the fraction becomes:
• 340/(340-90)=340/250=1.36

• A. is correct. The question tests your mastery of (a) algebraic manipulation and (b) the relationships among period, speed, frequency, and wavelength.
• Recalling that period is the reciprocal of frequency, and that
•      (speed)=(wavelength)(frequency)
•      (speed)=(wavelength)(1/period)
•      (speed)(period)=(wavelength)
• Wavelength is directly proportional to speed and frequency, so the ratio of WL₂ to WL₁ is simply S₂P₂:S₁P₁.

A. is correct. For any oscillatory motion that conforms to Hooke's law, restorative force and displacement are proportional. The question describes a swinging pendulum for which restorative force and displacement are not proportional. It is a simple logical deduction, therefore, that the pendulum does not conform to Hooke's law--for the reason as stated in the indicated correct choice, that "according to Hooke's law, restorative force and displacement are proportional."

• C. is correct. A linear equation as one that takes the form y=mx+b. Hooke's law represents a linear equation. Since Hooke's law, F=-k(x), represents such an equation, where m=-k and x=displacement, the graph drawn by the laboratory worker will be linear.
• For any linear equation, the slope has a magnitude equal to m, which in connection with Hooke's law, is -k, the spring constant. The graph drawn by this laboratory worker will be linear and will have a slope equal to the magnitude of the spring constant.

3x10⁸ meters per second (assigned the symbol c) and is the approximate speed at which electromagnetic radiation moves when traveling through space (a vacuum).

• The index of refraction is defined in terms of the degree to which it slows the travel of light. For any medium it is equal to the fraction
• (speed of light in vacuum)/(speed of light in medium)
• or n=c/v
• where n = refractive index of medium,
• c = speed of light in vacuum (3x10⁸ m/s)
• v = speed of light in medium
• Index of refraction is dimensionless (without units)

• visible light: electromagnetic radiation whose wavelength falls between 390x10^-9 meters and 700x10^-9 meters
• white light: a mixture of all wavelengths within the spectrum of visible light
• infrared light: electromagnetic radiation with wavelength somewhat above 700nm
• ultraviolet light: electromagnetic radiation falling somewhat below 390 nm

• The question requires algebraic application of the equation for refractive index..
• n=c/v = (3x10⁸m/s)/(2.26x10⁸m/s) = approx. 1.33

• The problems relies on the equation that defines refractive index.
• n=c/v
• 1.46=(3x10⁸m/s)/v
• v=approx. 2.5x10⁸m/s

• Reflection: the angles of incidence and reflection are equal.
• Refraction: the angles of incidence and refraction are related by Snell's law, which provides what for each such angle, there is equality between the product --
•    sin θ x index of refraction
• Thus, n₁sin θ₁=n₂sin θ₂
• where n₁ is the index of refraction for medium 1,
• n₂ is the index of refraction for medium 2,
• θ₁ is the angle of incidence, and
• θ₂ is the angle of refraction
• NOTE: The angles of incidence, refraction, and reflection are all measured by reference to an imaginary line oriented perpendicular to the medium interface.

• The 75 degree angle to which the problem refers is made between the reflected ray and the interface between the two media. It is NOT the angle of reflection. Rather the angle of reflection falls between the reflected ray and a line that runs perpendicular to the interface. The angle of reflection is complementary to the 75 degree angle mentioned in the problem:
• 90 degrees - 75 degrees = angle of reflection
•  = 15 degrees
• Since the angle of incidence and the angle of reflection are equal, the angle of incidence in this case is also 15 degrees.

• C. is correct. The 60° angle to which the problem refers does not constitute the angle of incidence, as it is formed by the media interface and the incident light ray. The angle of incidence is equal to the difference between that angle and 90°.
• The angle of incidence is, then, 90°-60°=30°. The sin of 30° is 0.5. Using Snell's law:
• n₁sinθ₁=n₂sinθ₂  R₁(0.5)=R₂sinθ₂
• [(0.5)(R₁)]/R₂=sinθ₂ 
• 1  x  R₁  =  sinθ₂      R₁   =  sinθ₂
• 2      R2                 2(R₂) 
•      θ₂=arcsin(R₁)/2(R₂)

• Total internal reflection refers to a ray of light that travels through a medium and encounters a second medium with a lower index of refraction than the first. If the angle between the incident ray and the line perpendicular to the interface exceeds the critical angle (θ_cr), all of the incident light will be reflected back into the first medium. The quantitative aspects of total internal reflection pertain to the evaluation of the critical angle for two interfaces, and are calculated with a modification of Snell's law:
• sin θ_cr=n_b/n_a
• where θ_cr = the critical angle that applies to the two media in question,
• n_b = the index of refraction for the second medium, and
• n_a = the index of refraction for the first medium

• f=(1/2)R_c
• where
• f= focal length, and
• R_c = mirror's radius of curvature
• Convex: the focal length extends into the mirror, and the focal point is said to be located behind it. (By convention, when the focal point is located behind a mirror, its associated focal length is given a negative sign.)
• Concave: the focal length is directed away from the mirror, and the focal point is said to be located in front of it. (By convention, when the focal point is located in front of a mirror, its associated focal length is assigned a positive sign.)

• The mirror is concave, and its focal point is located in front of the mirror, which gives it a positive sign.
• f=(1/2)R_c; R_c = 40cm
• f=(1/2)40cm = 20cm
• The focal length is +20cm.

• The focal length is directed into the mirror, and the focal point therefore carries a (-) sign. The absolute value of the focal length is equal to:
• f=(1/2)R_c; R_c = 162cm
• f=(1/2)162cm = 81cm
• The focal point is located -81 centimeters form the mirror.

• (1/f)=(1/o)+(1/i)
• where
•   f = the mirror's focal length,
•   o = the distance of the object from the mirror, and
•   i = the distance of the image from the mirror.
• Although a concave mirror has its focal point located in front of the mirror, it may produce images behind the mirror.

• Real: A mirror image that is located in front of the mirror -- which means it cannot be seen in the mirror but rather on a screen held in front of the mirror -- and it is designated with a positive (+) sign.
• Inverted: Any real image.
• Virtual: A mirror image that is located behind a mirror -- which means that it is visible in the mirror -- and it is designated with a negative (-) sign.
• Upright: Any virtual mirror image.

• Mirror Image Local ||  R/V   || Orientn  || Sign
• behind mirror        || virtual || upright  || (-)
• in front of mirror    ||  real   || inverted || (+)

• Find focal length
• f=(1/2)R_c; R_c = 80cm
• f=1/2(80cm) = 40cm
• Because the mirror is convex, f is negative.
• Focal length: =-40cm
• Find image location
• 1/f=1/o+1/i
• 1/-40cm = 1/20cm + 1/i
• 1/-40cm - 1/20cm = 1/i
• -1/40cm - 2/40cm = 1/i
• -3/40cm = 1/i
• Image location: =-(40cm/3) = appro.-13.33cm
• The negative sign indicates that the image is behind the mirror and thus virtual and upright.

• m=-i/o
• where
• m= magnification
• i= distance of image from mirror, and
• o= distance of object from mirror

• First, calculate the distance of the image from the mirror.
• Find focal length:
• f=(1/2)R_c; R_c = 60cm
• f=1/2(60cm) = 30cm
• Focal length: -30 cm (negative because it is a convex mirror)
• Find image distance
• 1/f = 1/o + 1/i
• 1/-30cm = 1/15cm + 1/i
• -1/30cm - 2/30cm = 1/i
• Image distance: =-30cm/3 = -10cm
• Find magnification;
• m=-i/o
• m=-(-10cm)/15cm = 10cm/15cm = 0.67
• Magnification: 0.67
• Since the object's true size is 20 cm,
• the mirror image will have a size equal to:
•   (0.67)(20cm) = 13.4 cm
• Note: A negative magnification would indicate an inverted image.

• Converging lenses produce the convergence of rays toward the lens's central axis; the focal point is that point at which parallel rays traversing the lens will come together though refraction. Diverging lenses produce the divergence of rays away from the central axis; the focal point represents that point at which parallel rays that have traversed the lens, and so have been subjected to divergence, will meet by hypothetical linear extensions.
•     Converging           Diverging
• 

• 
• where
•   f = focal length
•   n = refractive index of the lens material
•   R₁ = radius of curvature for side 1, and
•   R₂ = radius of curvature for side 2

• Imagine that light is incident from the right side, then side 1 is on the right, and R₁ = 5cm. Side 2 is on the left, and R₂ = 20cm. Since side 1 is on the right (and is convex), its center of curvature is on the left - the outgoing side - and it takes a positive sign, R₁ = +5cm. R₂, then, with center of curvature on the incoming side, takes a negative (-) sign and is equal to -20 centimeters. Application of the lensmaker's equation yields:
• 1/f=(n-1)[(1/R₁)-(1/R₂)]
•    =(1.5-1)[(1/5)-(1/-20)]
•    =(0.5)[(4/20)+(1/20)]
•    =(0.5)(5/20)
•    =(1/2)(1/4) = 1/8
• f=8.0cm

• With lenses, an image formed behind a lens is real and inverted, and one formed in front of a lens is virtual and upright.
•           ||   Type    ||    Image    || Focal point
•    For   || concave || real, invrt  || in front (+)
• Mirrors||  convex  ||virtual, uprt|| behind (-)
•    For   || divrgng  ||virtual, uprt|| in front (-)
• Lenses ||  cvrgng  ||real, invrtd || behind (+)

• The absolute value of the lens's focal length is 80cm; since the lens is divergent, the sign is negative (-). The focal length, therefore, is -80cm. To locate the image, apply the equation:
• 1/f=1/o+1/i
• -1/80cm=1/20cm+1/i
• -1/80cm-1/20cm=1/i
• -1/80cm-4/80cm=1/i
• -5/80cm=1/i
• i=-80/5cm = -16cm
• Image location: -16cm from the lens, which means it is 16 cm in front of the lens. Any image located in front of the lens is virtual and upright.
• Image size is = object size x magnification
• m=-i/o = -(-16cm)/20cm = 0.8
• Image size: (40cm)(0.8) = 32cm

• Focal power expresses the degree to which a lens imposes a convergence or divergence on the light rays that traverse it. The converging lens of relatively greater power produces greater convergence than does one of relatively lesser power; the diverging lens of relatively greater power produces greater divergence than does one of relatively lesser power.
• It is expressed in the unit diopter.
• It is equal to the reciprocal of the focal length:
•   Focal Power (P_f)=1/f
• Power of converging lens carries positive sign.
• Power of diverging lens carries negative sign.

• The illustration reveals a convergent lens with a focal length of +22cm. Expressed in meters, the focal length is +0.22meters, and the focal power is:
• 1/0.22 diopters
• = approx. 4.5 diopters

• C. is correct. The patient is unable to focus objects in the distance, suggesting that as parallel light rays traverse the refractive components of his eye, they experience such convergence as produces a focal point anterior to the retina. Either one or both of his eyes has excessive focal power or his retina is located too far back.
• That the patient clearly sees objects located near to him tends to confirm the physician's hypothesis, since, as an object comes closer to a lens, the rays that emanate from it show increased divergence in relation to one another. For such rays, the focal point would be moved backward and might focus on the retina as they should.
• If the physician's hypothesis is correct--that in relation to the retina's location, parallel rays of light are focused too far forward--some improvement should accompany the placement of a concave lens in front of the eye. The lens would diverge the incoming light rays before they reached the eye, and would present the eye with rays that, in their divergence, resemble rays that come from objects situated close to it. If the placement of a concave lens brings improvement, the hypothesis is supported.

A. is correct. Using the rules for right-angled triangles, tan θ₂ = 1000/1700 = 1/1.7. From knowledge, tan 30^o = 1/(√3), which is approximately equal to 1/1.7. Therefore, θ₂ is approximately 30^o.

• A. is correct.  n = c/v, n is inversely proportional to v
• c=vn, or c=v₁n₁ and c=v₂n₂,
• ∴ v₁n₁ = v₂n₂ = constant
• Thus v_redn_red = v_violetn_violet = constant 
• ∴ given ↓v_violet then ↑n_violet
• and since the value of v_violet is lower than the value of v_red, then the increase for n_violet will be greater than the value of n_red.

B. is correct. Using the derivation of Snell's law which refers to the critical angle, sin θ_c = n₂/n₁= n_air/n_water (see question text), where "θ_c" is the unknown critical angle. Thus sin θ_c = 1.00/1.33 ≈ 1/(4/3) = 3/4 = 0.75.

• D. is correct. First: sin θ_c; = n₂/n₁ = n_air/n_water
• Now: sin θ_c2 = (n_air + x)/n_water
• Thus we increased the numerator by some value "x" which means sin θ_c2 > sin θ_c, which also means θ_c2 > θ_c.

C. is correct. Both velocity and wavelength (different colors) change (recall velocity and wavelength are directly proportional). One should also know that the intensity (= the rate of energy propagation through space) of a light ray decreases when it passes through a medium other than air (its velocity decreases in the medium and this leads to an overall loss of energy).

C. is correct. Since the 200 N block is not in motion, the value of the frictional force is equal and opposite in direction to the applied force, that is, 10 N.

D. is correct. From the equation F = kq1q2/r2, if the distance r is doubled then (2r)2 = 4r2, which means 4 x r2 in the denominator. Thus the original force is quartered (=decreased by a factor of 4).

• D. is correct. Let the number of decibels of sound Y be dB_Y = 10 log (I/I_o). If the intensity 'I' of sound X increases by a factor of 1000 then:
• dBX = 10 log (1000 I/I_o) = 10 log (10)³ + 10 log (I/I_o) = 30 + dB_Y.

• A. is correct. A simple torque force problem: let the fulcrum (= the center of gravity of the bar) be the pivot point (draw a vector diagram):
• ∑L; = CCW - CW 
•  =(100 kg)(g)(0.5 m) - (75 kg)(g)(x) = 0
• Thus (100 kg)(g)(0.5 m) = (75 kg)(g)(x)
• g cancels, manipulate to get: x = 50/75 = 2/3 m = 0.67 m (CW is to the right of the fulcrum).

D. is correct. The external force acting on the cars is tiny compared to the large forces of the colliding vehicles (i.e. friction is negligible), so momentum is conserved (PHY 4.3). The total kinetic energy before collision is not equal to the total kinetic energy after the collision, thus it is not conserved and the collision is not completely elastic. It must be remembered that during a collision energy may change form (i.e. kinetic, heat, sound, etc.), but total energy must be conserved.

• B. is correct. Using the principle of Conservation of Momentum:
• Momentum before collision = Momentum after collision
• (1200 kg)(7.5 m/s) + (8000 kg)(3.0 m/s) = (1200 kg)(3.0 m/s) + (8000 kg)(x m/s)
• Divide through by 400:
• (3)(7.5) + (20)(3) = (3)(3) + 20x
• Isolate x: (7.5 + 20 - 3)3/20 = x
• Thus x = (24.5)(3/20) = 73.5/20 = 3 13.5/20 ≈ 3 7/10 = 3.7 m/s.
• {Recall: momentum is a vector; in this problem the sign for the velocities are always positive because we are told that the vehicles are always moving in the same (i.e. northerly) direction}

• C. is correct. F = f < f_max; KW: tries
• This question is fun! We begin with the fact that the maximum frictional force that can be exerted occurs when the car is just about to move, and is equal to the product of the coefficient of friction (static) and the normal force on the car. Since the car is not moving perpendicular to the surface, using Newton's Second Law (F = ma), since the acceleration 'a' when the car is about to move is zero, the normal force is equal to the weight of the car (i.e. ∑F = N - Mg = ma = 0; thus N = Mg).
• Therefore, the maximum frictional force = (1/3)(1200 kg)(10 m/s²)= 4000 N, approximately since we estimated 'g'. Since the force the car exerts (300 N) is less than the maximum frictional force on the car (4000 N), the car is not in motion. For an object not in motion, the sum of forces must be equal to zero (Newton # II !). Because the exerted force is 300 N, the actual frictional force must be 300 N in the opposite direction.

B. is correct. The normal force on an object always acts perpendicular to the surface, in this case, the road. Friction depends on the normal force. Since only a component of the weight of the car acts perpendicular to the hill, the value of the normal force decreases. Since the coefficient of friction remains the same, the value of the maximum frictional force decreases. {Once an object is on an incline: N < weight of the object as determined by the cosine of the angle of the incline}

• A. is correct. The potential energy (P.E.) = mass (m) x gravitational acceleration (g) x perpendicular height (h).
• "h" is given by 40sin30^o = (40)(1/2) = 20 m.
• The weight "m x g" is given already as
• 12000 N. Thus (20)(12000) = 240000 J.

• C. is correct. Dimensional analysis is cool because it means you don't need to memorize an equation (!), you just have to define the terms and follow the units.
• Area of collector = (20 m)(20 m) = 400 m²
• Incident solar energy = 1600 W/m² = 1600 J/sm2
• Therefore, energy received per second by collector = 1600 J/sm² x 400 m² = 640000 J/s
• Rate of flow of water = 4 L/s = 4000 cm3/s 
• Mass of water going through collector in one second = (4000 cm³)(1 g/cm³) = 4000 g
• From Q = mcΔt: 640000 J = (4000 g)(4.2 J/goC)(Δt)
• Thus Δt = 38.1 ^oC
• {The answer could be quickly estimated by approximating 'c' as 4 and dividing through by 16000; the answer obtained would be 40 ^oC which is sufficiently accurate for this problem}

B. is correct. The specific heat capacity is the energy required to raise the temperature of a unit of mass of a substance by one degree. The smaller the specific heat capacity, the less energy is required to raise the temperature of that substance by a certain amount.

• D. is correct. There are two parts to this problem. We begin by calculating the impact velocity of the 0.2 kg ball using the equation for speed V at any time t:
• V = V_o + at, where V_o is the original speed. Since the ball starts at rest we get:
• V = 0 m/s + (5 m/s²)(5 s) = 25 m/s
• Using the principle of Conservation of Momentum and momentum = mass x velocity, we get:
• Total momentum before collision = total momentum after collision
• (0.2 kg)(25 m/s) + (0.5 kg)(0 m/s)
• = (0.2 kg)(0 m/s) + (0.5 kg)(x m/s)
• Isolate x: x = 10.0 m/s

• A. is correct. F = mg = kx; k leading to W
• At equilibrium the force of the weight (W = mg) will be equal to the spring force (Fs = -kx), thus: kx = mg, k = mg/x
• = (1 kg x 10 m/s²)/(0.5 m) = 20 kg/s²
• Work done (spring) = 1/2 kx² 
• = 1/2 (20)(0.5)² 
• = 1/2 (20)(1/4)
• = (20)/(8) = 5/2 = 2.5 J.
• {Note F = mg, g is estimated as 10 m/s²}