Review: work and energy
Pull the work and energy unit together. Read the summary, then try each question before opening its solution. The lesson number next to each problem tells you where to look if you get stuck.
Key ideas of the unit
| Idea | Equation | Lesson |
|---|---|---|
| Work by a constant force | \(W = Fd\cos\theta\); positive, negative or zero | 9 |
| Kinetic energy | \(KE = \tfrac12 mv^2\) | 10 |
| Work-kinetic energy theorem | \(W_{\text{net}} = \Delta KE\) | 10 |
| Gravitational potential energy | \(PE_g = mgh\), from a chosen zero level | 11 |
| Elastic potential energy | \(PE_{\text{elastic}} = \tfrac12 kx^2\) | 11 |
| Conservation of mechanical energy | \(KE_i + PE_i = KE_f + PE_f\) (no friction) | 12 |
| Power | \(P = W/\Delta t = Fv\) | 13 |
Concept questions
Question 1
In which case is work done on the object?
- A weightlifter holds a barbell motionless overhead.
- A waiter carries a tray across a room at constant speed.
- A student lifts a backpack from the floor onto a desk.
- A person pushes hard on a parked truck that does not move.
Show solution
Only lifting the backpack combines a force with a displacement along it. In (a) and (d) nothing moves; in (b) the upward force is perpendicular to the motion.
Question 2
A ball is thrown straight up. On the way up, what happens to its kinetic and potential energy? (Ignore air resistance.)
- Both increase.
- Kinetic energy decreases and potential energy increases by the same amount.
- Kinetic energy stays the same and potential energy increases.
- Both decrease.
Show solution
Gravity does negative work on the rising ball, so it slows down, while its height and \(PE_g\) grow. With no air resistance the total stays constant.
Question 3
A car brakes with the same force from 15 m/s and from 30 m/s. How does the stopping distance from 30 m/s compare?
- The same
- Twice as far
- Three times as far
- Four times as far
Show solution
\(Fd = \tfrac12 mv^2\), so \(d\) is proportional to \(v^2\). Doubling the speed quadruples the stopping distance.
Question 4
Two cranes lift identical loads to the same height. Crane A takes 20 s and crane B takes 40 s. Which statement is true?
- They do the same work, and A has twice the power.
- A does twice the work, and they have the same power.
- B does twice the work, and A has twice the power.
- They do the same work and have the same power.
Show solution
Same load, same height: same work \(mgh\). A does it in half the time, so \(P = W/\Delta t\) is twice as large.
Problems
Problem 1 · Lesson 9
A child pulls a toy wagon 12.0 m with a force of 75.0 N directed 30.0° above the horizontal. How much work does the child do?
Show solution
\(W = (75.0)(12.0)\cos 30.0^\circ = 779\ \text{J}\)
Problem 2 · Lesson 10
A 60.0 g tennis ball is served at 50.0 m/s. What is its kinetic energy?
Show solution
\(KE = \tfrac12 (0.0600)(50.0)^2 = 75.0\ \text{J}\)
Problem 3 · Lesson 10
A 1000 kg car traveling at 20.0 m/s brakes to a stop in 40.0 m. What is the average braking force?
Show solution
\(W_{\text{net}} = \Delta KE\): \(-F(40.0) = 0 - \tfrac12(1000)(20.0)^2 = -2.00 \times 10^5\ \text{J}\), so \(F = 5.00 \times 10^3\ \text{N}\).
Problem 4 · Lesson 11
A door spring with \(k = 800\ \text{N/m}\) is compressed 5.0 cm. How much elastic energy does it store?
Show solution
\(PE = \tfrac12 (800)(0.050)^2 = 1.0\ \text{J}\)
Problem 5 · Lesson 12
A skateboarder starts from rest at the top of a frictionless ramp 4.00 m high. How fast is he moving when he is 1.50 m above the bottom?
Show solution
He has dropped \(4.00 - 1.50 = 2.50\ \text{m}\): \(v = \sqrt{2(9.81)(2.50)} = 7.00\ \text{m/s}\). At the bottom he would reach \(\sqrt{2(9.81)(4.00)} = 8.86\ \text{m/s}\).
Problem 6 · Lesson 12
A 50.0 kg skier starts from rest and descends a vertical drop of 15.0 m, reaching 14.0 m/s at the bottom. How much mechanical energy was lost to friction and air resistance?
Show solution
\(ME_i = (50.0)(9.81)(15.0) = 7358\ \text{J}\); \(ME_f = \tfrac12(50.0)(14.0)^2 = 4900\ \text{J}\). Lost: \(7358 - 4900 = 2.46 \times 10^3\ \text{J}\).
Problem 7 · Lesson 13
A crane lifts a 500 kg beam 20.0 m at constant speed in 25.0 s. What is its power output?
Show solution
\(P = \dfrac{mgh}{\Delta t} = \dfrac{(500)(9.81)(20.0)}{25.0} = 3.92 \times 10^3\ \text{W}\)
Multi-step problems
Problem 8 · Lessons 10 to 12
A spring with \(k = 400\ \text{N/m}\) is compressed 0.200 m and released, pushing a 2.00 kg block across a level floor. The block leaves the spring and slides on. The coefficient of kinetic friction is 0.250, and you can ignore friction while the spring is pushing. How far does the block slide after leaving the spring?
- Energy stored in the spring. \(PE = \tfrac12 (400)(0.200)^2 = 8.00\ \text{J}\). This becomes the block's kinetic energy.
- Friction force. \(F_k = \mu_k mg = 0.250 \times 2.00 \times 9.81 = 4.91\ \text{N}\).
- Work-kinetic energy theorem. Friction removes all 8.00 J: \(F_k d = 8.00\ \text{J}\), so $$d = \frac{8.00}{4.91} = 1.63\ \text{m}$$
The block slides 1.63 m.
Problem 9 · Lesson 12
A 30.0 kg sled moves at 3.00 m/s at the top of a hill 8.00 m high. On the way down, friction does \(-1200\ \text{J}\) of work on it. How fast is it moving at the bottom?
- Energy at the top. \(KE_i = \tfrac12(30.0)(3.00)^2 = 135\ \text{J}\) and \(PE_i = (30.0)(9.81)(8.00) = 2354\ \text{J}\).
- Energy at the bottom. Start with the total and remove what friction took: \(KE_f = 135 + 2354 - 1200 = 1289\ \text{J}\).
- Speed. $$v_f = \sqrt{\frac{2(1289)}{30.0}} = 9.27\ \text{m/s}$$
\(v_f = 9.27\ \text{m/s}\)
Ready for the next unit?
- If these went well, continue with Momentum and impulse.
- If energy conservation problems were hard, redo the worked examples in Conservation of mechanical energy with the steps hidden.
Finished?
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