Physics 1100 Work & Energy Solutions

Physics 1100: Work & Energy Solutions

In the diagram below, calculate the work done if

(a) F = 15.0 N, θ = 15°, and Δx = 2.50 m,

(b) F = 25.0 N, θ = 75°, and Δx = 12.0 m,

For constant forces, work is defined by W = FΔxcos(θ).

(a) W = 36.2 J
(b) W = 77.6 J
(c) W = -38.9 J

In the diagram below, a rope with tension T = 150 N pulls a 15.0-kg block 3.0 m up an incline (θ = 25.0°). The coefficient of kinetic friction is μ_k = 0.20. Find the work done by each force acting on the block.

To find the work done by a force, we need to know the magnitude of the force and the angle it makes with the displacement. To find forces, we draw a FBD and use Newton's Second Law.

i	j
F_x = ma_x	F_y = ma_y
T - f_k - mgsin(θ) = ma	N - mgcos(θ) = 0

The second equation informs us that N = mgcos(θ). We know f_k = μ_kN = μ_kmgcos(θ).

*Force*	*Force* (N)	θ	W = FΔxcos(θ)(J)
Tension	150	0	450
Weight	147.15	θ + π/2	-187
Normal	133.36	π/2	0
Friction	26.67	π	-80

A winch lifts a 150 kg crate 3.0 m upwards with an acceleration of 0.50 m/s². How much work is done by the winch? How much work is done by gravity?

To find the work done by a force, we need to know the magnitude of the force and the angle it makes with the displacement. To find forces, we draw a FBD and use Newton's Second Law.

F_y = ma_y

T - mg = ma

The work done by the winch is the work done by tension. The work done by gravity is the work done by the object's weight. Since we know m and g, we find T = mg + ma = 1546.5 N. The work done by tension is W_tension = TΔycos(0) = 4.64 × 10³ J. The work done by gravity is W_gravity = mgΔycos(π) = -4.41 × 10³ J.

What work does a baseball bat do on a baseball of mass 0.325 kg which has an initial speed forward of 36 m/s and a final speed of 27 m/s backwards. Assume motion is linear and horizontal. The work done by the bat is a non-conservative force.

Since we are asked for the work done and have a change in speed, we make use of the generalized Work-Energy Theorem. Since the height of the ball does not change, there is only a change in kinetic energy.

W_NC = E = K_f - K_i = ½m[(v_f)²- (v₀)²] = ½(0.325kg)[(-27 m/s)²- (36 m/s)²] = -92.1 J .

This is the work done on the ball by the bat. It's not a good hit as the ball slowed down. The batter decreased the energy of the ball. Perhaps he was trying for a bunt!

How much work must be done to stop a 2000-kg car travelling at 60 km/h in 15.0 m? What was the average breaking force?

Since we are asked for the work done and have a change in speed, we make use of the generalized Work-Energy Theorem. Since the height of the car does not change, there is only a change in kinetic energy. First converting the initial velocity into SI

60 km/h = 60 km/h × (1000 m)/(1 km) × (1 h)/(3600 s) = 16.67 m/s .

Therefore,

W_brake = E = K_f - K_i = ½m[(v_f)²-(v₀)²] = ½(2000 kg)[(0)²-(16.67 m/s)²] = 2.77810⁵ J .

Now the force doing this work, f_brake, is related to the work by W_brake = f_brakexcos(θ). Since the force is slowing the car down, θ = 180°, cos(180°) = -1, and

f_brake = -W_brake / Δx = -(2.778 × 10⁵ J )/(15.0 m) = 1.85 × 10⁴ N .

In the diagram below, determine the speed of the block at each point. Assume no friction. The mass of the block is 10.0 kg.

Since the problem involves a change in height and speed, we make use of the generalized Work-Energy Theorem,

W_NC = E = P_f - P_i + Kf - K_i = mg(h_f - h_i) + ½m[(v_f)²-(v_i)²] .

Since there is no mention of friction, W_NC = 0. Our equation therefore simplifies to

mg(h_f - h_i) + ½m[(v_f)²-(v_i)²] = 0 ,

or more simply

mgh_f + ½m(v_f)² = mgh_i + ½m(v_i)² .

We can divide through by m, and since we know h_f,h_i, and v_i, we can rearrange the above to find v_f

(v_f)² = (v_i)² + 2g(h_i - h_f) .

For the given values, we find

	h_f (m)	v_i (m/s)
1	15	5
2	10	11.1
3	5	14.9
4	0	17.9

A 2.0-kg rock is thrown with initial speed of 9.8 m/s at an unknown angle. The speed of the rock at the top of the parabola is 2.1 m/s. How high does it go? Assume no air resistance.

Since the problem involves a change in height and speed, we make use of the generalized Work-Energy Theorem,

W_NC = E = P_f - P_i + Kf - K_i = mg(h_f - h_i) + ½m[(v_f)²-(v_i)²] .

Since we are told that there is no air resistance, W_NC = 0. Our equation therefore simplifies to

mg(h_f - h_i) + ½m[(v_f)²-(v_i)²] = 0 ,

or more simply

mgh_f + ½m(v_f)² = mgh_i + ½m(v_i)² .

We can divide through by mg, and since we know h_i, v_i,and v_f, we can rearrange the above to find h_f

h_f = [(v_i)²- (v_f)²]/2g = [(9.8 m/s)² - (2.1 m/s)²]/(29.81 m/s²) = 4.67 m .

The rock reaches 4.67 m up into the air.

In the diagram below, a 5.00-kg block slides from rest at a height of h₁ = 1.75 m down to a horizontal surface where it passes over a 2.00-m rough patch. The rough patch has a coefficient of kinetic friction μ_k = 0.25. What height, h₂, does the block reach on the incline?

Since the problem involves a change of height and speed, we make use of the Generalized Work-Energy Theorem. Since the block's initial and final speeds are zero, we have

W_NC = E = U_f - U_i = mgh₂- mgh₁ . (1)

The nonconservative force in this problem is friction. To find the work done by friction, we need to know the friction. To find friction, a force, we draw a FBD at the rough surface and use Newton's Second Law.

i	j
F_x = ma_x	F_y = ma_y
- f_k= -ma	N - mg = 0

The second equation gives N = mg and we know f_k = μ_kN, so f_k = μ_kmg. Therefore, the work done by friction is W_friction = -f_kΔx = -μ_kmgΔx. Putting this into equation (1) yields

-μ_kmgΔx = mgh₂- mgh₁ .

Solving for h₂, we find

h₂ = h₁ - μ_kΔx = 1.25 m .

In the diagram below, a 5.00-kg block slides from rest at a height of h₁ = 1.75 m down to a smooth horizontal surface until it encounters a rough incline. The incline has a coefficient of kinetic friction μ_k = 0.25. What height, h₂, does the block reach on the θ = 30.0° incline?

Since the problem involves a change of height and speed, we make use of the Generalized Work-Energy Theorem. Since the block's initial and final speeds are zero, we have

W_NC = E = U_f - U_i = mgh₂- mgh₁ . (1)

i	j
F_x = ma_x	F_y = ma_y
-f_k - mgsin(θ) = -ma	N - mgcos(θ) = 0

The second equation gives N = mgcos(θ) and we know f_k = μ_kN, so f_k = μ_kmgcos(θ). Therefore, the work done by friction is W_friction = -f_kΔx = -μ_kmgcos(θ)Δx. Putting this into equation (1) yields

-μ_kmgcos(θ)Δx = mgh₂- mgh₁ .

A little trigonometry shows that Δx is related to h₂ by Δx = h₂ / sin(θ). Putting this into the above equation yields

-μ_kcos(θ)[ h₂ / sin(θ)] = h₂ - h₁ .

Solving for h₂, we find

h₂ = h₁ / [1 + μ_k/tan(θ)] = 1.22 m .

In the figure below, a block of mass 5.0 kg starts at point A with a speed of 15.0 m/s on a flat frictionless surface. At point B, it encounters an incline with coefficient of kinetic friction μ_k = 0.15. The block makes it up the incline to a second flat frictionless surface. What is the work done by friction? What is the velocity of the block at point C? The incline is 2.2 m long at an angle θ = 15°.

The problem involves a change in height and speed, so we apply the generalized Work-Energy Theorem.

W_NC = E = (K_f - K_i) + (U_f - U_i) = ½m(v_C)² - ½m(v_A)² + mgh . (1)

Here the nonconservative force is friction, so W_NC = W_f. To find friction, a force, we draw a FBD and use Newton's Second Law.

i	j
F_x = ma_x	F_y = ma_y
-f_k - mgsin(θ) = -ma	N - mgcos(θ) = 0

The second equation gives N = mgcos(θ) and we know f_k = μ_kN, so f_k = μ_kmgcos(θ). Therefore, the work done by friction is

W_f = -f_kΔx = -μ_kmgcos(θ)Δx = -(.15)(5 kg)(9.81 m/s²)cos(15°)(2.2 m) = -15.635 J.

Note from the diagram, that the height h is related to the length of the incline by h = Δxsin(θ). Putting both results into equation (1) yields

W_f = ½m(v_C)² - ½m(v_A)² + mg[Δxsin(θ)] .

Solving for v_C yields

v_C = [2W_f/m - 2gΔxsin(θ) + (v_A)²]^½ = 14.4 m/s .

Two blocks are connected by a string hung over a frictionless massless pulley. Block A has mass M_A and block B has mass M_B. Initially the blocks are held at rest before being allowed to move. How fast will block B be moving when it has risen a distance h?

Again we have a change in height and speed, so we apply the Work-Energy Theorem

W_NC = (K_f - K_i) + (U_f - U_i).

We are told that there is no friction so W_NC = 0.

The difference between this and earlier problems is that we are dealing with two objects. For each object there is an external force the tension T in the string. However the work done by the tension in each case is equal, since the distance each block moves is the same, but opposite. (Check this!) So for the system, energy is transferred from one block to the other. We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

0 = [½M_Bv_f² + M_Bgh] + [½M_Av_f² – M_Agh]

Note that the two blocks are connected by a string so the final speed of each is the same. Also if block B moves up h block A drops h. Thus our equation becomes

0 = ½ (M_A + M_B)v_f² – (M_A – M_B)gh.

When we solve this, we find

Two blocks are connected by a string hung over a frictionless massless pulley. Block A has mass M_A and is on a table top. Block B has mass M_B and is hanging in the air. Initially the blocks are held at rest. The coefficients of friction between block A and the tabletop are μ_S and μ_K.
(a) B is allowed to fall. How fast will block B be moving when it has fallen distance h?
(b) Block A is pulled to the left by a horizontal force F for a distance L. How fast will block B be moving?

(a) Again we have a change in height and speed, so we apply the Work-Energy Theorem

W_NC = (K_f - K_i) + (U_f - U_i).

We are told that there is friction so we need to determine W_NC = W_friction. Friction does negative work, takes energy out of the system, since it is opposite to the movement of block A. To find friction, a force, we draw a FBD of block A and use Newton’s Second Law.

		i		j
		ΣF_x = ma_x		ΣF_y = ma_y
		T - f_k= M_Aa		N - M_Ag = 0

The second equation gives N = M_Ag and we know f_k = μ_kN, so f_k = μ_kM_Ag. Therefore, the work done by friction is W_friction = -f_kΔx = -μ_kM_Agh since block A will move as far as block B will drop.

For the pair of block, the tension T in the string, is internal and does not net work. So for the system, energy is transferred from one block to the other. We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

-μ_kM_Agh = ½M_Av_f² + [½M_Bv_f² – M_Agh]

Note that the two blocks are connected by a string so the final speed of each is the same. Thus our equation becomes

M_Bgh – μ_kM_Agh = ½(M_A + M_B)v_f².

When we solve this, we find

(b) Again we have a change in height and speed, so we apply the Work-Energy Theorem

W_NC = (K_f - K_i) + (U_f - U_i).

We are told that there is friction, and the work done by friction is still W_friction = -f_kΔx = -μ_kM_AgL since block A moves L not h. Because of the string block B rises L and both blocks will have the same speed. However there is an extra external force F which in the same direction as the motion of block A. It does positive work adding to the energy of the system.

We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

FL –μ_kM_Agh = ½M_Av_f² + [½M_Bv_f² + M_AgL]

Thus our equation becomes

FL – M_Bgh – μ_kM_Agh = ½(M_A + M_B)v_f² .

When we solve this, we find

Two blocks are connected by a sting slung over a pulley as shown in the diagram below. The hanging block is allowed to drop. How fast will it be moving when it hits the ground? The block on the incline has mass M_A = 2.50 kg. The hanging block has mass M_B = 1.50 kg. The incline makes and angle θ = 30° with horizontal. Ignore friction.

Again we have a change in height and speed, so we apply the Work-Energy Theorem

W_NC = (K_f - K_i) + (U_f - U_i).

We are told to ignore friction so W_NC = 0.

0 =	(½M_BV_Bf² - 0) + (M_Bg(0) - M_Bg(1.0m))
	+ (½M_AV_Af² - 0) + (M_Ag(h_Af - h_Ai))

Now the two blocks are connected by a string so the final speed of each is the same, V_Bf = V_Af = V_f. Next the block moves 1.0 m up the 30° degree incline, so h_Af - h_Ai = (1.0 m)sin(30°). Thus our equation becomes

0 = ½M_BV_f² + ½M_AV_f² - M_Bg(1.0m) + M_Ag(1.0m)sin(30°) .

When we solve this we find

V_f = {2(9.81)[(1.50)(1.0m) - (2.50)(1.0m)sin(30°)]/(1.50 + 2.50)}^½ = 1.338 m/s .

In the diagram below, what is the minimum height that the skier must start from to successfully make it around the loop. Assume (a) no friction, and (b) that friction does -3.0 × 10³ J of work on the skier. The radius is 5.00 m and the skier has mass 65.0 kg.

The problem involves a change of height and speed, so that suggests that we use the generalized Work-Energy Equation. However, the skier also travels in a circle, which suggests a centripetal acceleration problem. Centripetal acceleration problems are solved by drawing a free-body diagram (FBD) and applying Newton's Second Law. Let's do this first.

At the top of the inside of the loop, the centripetal acceleration acts straight down as does the normal force and the weight.

F_y = ma_y

-N - mg = -m(v_f)²/r

The skier will lose contact with the inside of the loop when N goes to zero. This fact and our equation, let's us find a minimum value of v_f,

v_f = [gr]^½ = [(9.81 m/s²)(5.00 m)]^½ = 7.004 m/s .

Now we consider the work energy portion of the problem.

The Work-Energy formula may be rewritten as

mgh_f + ½m(v_f)² = mgh_i + ½m(v_i)² .

We know v_i = 0, we see from the diagram that h_f = 2r, and v_f = [gr]^½from our earlier work, so we rearrange the above equation to find h_i

h_i = h_f + ½(v_f)²/g = 2r + ½r = (5/2)r = 12.5 m .

If the trip is frictionless, the hill needs to be at least 12.5-m tall if the skier is to make it around the loop safely.

Since there are non-conservative forces, the generalized Work-Energy equation for this case is

W_NC = [mgh_f + ½m(v_f)²] - [mgh_i + ½m(v_i)²] .

We are told W_NC = -3000 J, so we rearrange the equation to find that h_i is,

h_i = {[mgh_f + ½m(v_f)²] - W_NC }/mg = (5/2)r - W_NC/mg .

Using the given data,

h_i = 12.5 m - (-3000 J)/(65.0 kg)(9.81 m/s²) = 17.2 m .

With this much friction, the hill needs to be at least 17.2-m tall if the skier is to make it around the loop safely.

Tarzan, Lord of Apes, is swinging through the jungle. In the diagram below, Tarzan is standing at point A on a tree branch h₁= 22.0 m above the floor of the jungle. Tarzan is holding one end of a vine which is attached to a branch on a second tree. The vine is L = 21.0 m long. When Tarzan swings on the vine, his path is in an arc of a circle. At the bottom of his swing he is at point B, 13.0 m above the ground . Ignore Tarzan's height. Tarzan has a mass of 90.0 kg. The vine does not stretch and has negligible mass.
(a) Why does the tension in the vine do no work?
(b) What will be his speed at point B?
(c)What will be the tension in the rope at point B?

The problem involves a change in height and speed, so we apply the generalized Work-Energy Theorem.

W_NC = E = (K_f - K_i) + (U_f - U_i) = ½m(v_B)² - mgh . (1) (a) Here the only possible nonconservative force is friction, so W_NC = W_T. The definition of work is W = Fxcos, but in this problem the tension is along a radius and is thus always at 90 to the displacement. As a result, W_T = 0. Thus we have

0 = ½m(v_B)² - mgh .

(b) To find the speed at point B, we need to know h, the distance Tarzan dropped. Examining the question, we see that h = h₁ - h₂ = 22.0 m - 13.0 m = 9.0 m. Rearranging our equation, we find

v_B = [2gh]^½ = [2(9.81 m/s²)(9 m)]^½ = 13.29 m/s .

(c) Tension is a force. To find a force we need to draw a FBD and apply Newton's Second Law. Since Tarzan is swinging in a circle, we are dealing with centripetal acceleration.

F_y = ma_y

T - mg = mv²/L

Solving for T,

T = mg + mv²/L = (90.0 kg)[ 9.81 m/s² + (13.288 m/s)²/(21.0 m)] = 1.64 10³ N.

A 0.200-kg block slides down the track and horizontally off a table as shown in the diagram below.
(a) Assuming that friction is negligible, how far from the table does the block land?
(b) The block only land 1.20 m away. How much work was done by friction and other non-conservative forces?

The first part of the problem involves a change in height and speed, so we can use the Work-Energy Theorem there. When the block leaves the surface it becomes a projectile.

(a) Applying the Work-Energy Theorem and assuming that the initial velocity of the block is zero.

W_NC = ΔE = (K_f - K_i) + (U_f - U_i) = ½m(v_B)² - 0 + mg(h_f - h_i) .

The mass m cancels out and we find

v_B = [-2g(h_f - h_i)]^½ = [-2(9.81)(1.0 - 1.5)]^½ = 3.3121 m/s .

Now this velocity is the initial velocity for the projectile.

i	j
v_0x = 3.3121 m/s	v_0y = 0 m/s (horizontal flight)
a_{x = 0 m/s²}	a_y = -9.81 m/s²
Δx = ?	Δy = -1.0 m
----- t (common) -----

From the j information we can find the time that the block is in the air using Δy = v_0yt + ½a_yt². This becomes -1.0 m = ½(-9.81 m/s²)t² or t = ±0.4515 s. We need the positive, forward in time, solution. We then find Δx using

Δx = v_0xt + ½a_xt² = (3.3121 m/s)(0.4515 s) = 1.4955 m.

The block lands 1.50 m from the edge of the table.

(b) If the block only lands 1.20 m away, then is velocity must have been v_0x = (1.20 m)/(0.4515 s) = 2.6578 m/s .

This is also the velocity at the bottom of the slide. To find W_NC we again use the Work-Energy Theorem.

W_NC = (K_f - K_i) + (U_f - U_i) = ½m(v_0x)² - 0 + mg(h_f - h_i) .

So the work done by non-conservative forces is

W_NC = ½(0.200)(2.6578)² + (0.200)g(1.0 - 1.5) = -0.2746 J.

A 50-hp engine is used to lift heavy loads at a worksite. It is used to lift a load of bricks weighing 2000 N to the top of a new building 35.0 m above ground. How long does it take for the load to get to the top?

We are given the power of the engine

P = 50 hp × (740 W/hp) = 37,000 W.

Power is defined as work done per given time, P = W/t. The time t is what we are asked for. Work done is force times distance, here W = 2000 N × 35 m = 70,000 J.

So the time needed is

t = W/P = 70,000 J / 37,000 W = 1.89 seconds.

Note however that rated power is seldom the same as the actual power that does useful work.

A 5.0 MW generating station is situated at a 22-m high dam. The energy used to generate the electricity comes from the loss in potential energy of the water as it falls the height of the dam. What is the minimum amount of water going through the dam every day?

We are given the power (50,000,000 W) and power is defined as work done per given time, P = W/t. The time t we are given is one day. We are told that the work done equals the loss in potential energy of the water falling from the top of the dam, so W = mgh where h = 22.0 m. Thus the amount of water, i.e. its mass, is found from P = mgh / t or

m = Pt/gh = (5 × 106)(1 d × 24 h/d × 3600 s/h) / (9.81)(22.0) = 2.00 × 109 kg.

What power is required to pull a 5.0-kg block at a steady speed of 1.25 m/s? The coefficient of friction is 0.30.

The power required to move the block at constant speed is P = Fv. We are given v, the speed of the block. To get F, a force, we draw a FBD and apply Newton's Second Law,

i	j
F_x = ma_x	F_y = ma_y
F - f_k= 0	N - mg = 0

The second equation gives N = mg and we know f_k = μ_kN, so f_k = μ_kmg. Therefore, the applied force is F = μ_kmg. Thus the power is

P = μ_kmgv = (0.3)(5 kg)(9.81 m/s²)(1.25 m/s) = 18.4 Watts.

A 7000-W engine is propelling a speedboat at 30 km/h. What force is the engine exerting on the speedboat? What force and how much power is water resistance exerting on the speedboat?

First we convert the velocity to SI units,

30 km/h × (1000 m)/km × (1 h)/(3600 s) = 8.333 m/s .

We know P = Fv, so

F = P/v = 7000 W / 8.333 m/s = 840 N .

By Newton's Third Law, the water is exerting 840 N in the reverse direction. It is also removing 7000 W of power which is going into increasing the kinetic energy of the water.

Questions? mike.coombes@kwantlen.ca