This diagnostic is for the algebra-based Physics 1 course, Unit 3 (Work, Energy, and Power). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.
These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator in degree mode and g = 9.8 m/s² (9.8 N/kg). Ignore air resistance, and treat springs as ideal.
Question 1 (multiple choice · 3.1)
A 0.80 kg ball moves east at 3.0 m/s. After a kick it moves west at 5.0 m/s. What is the change in the ball’s kinetic energy?
- (A) 26 J
- (B) 6.4 J
- (C) −14 J
- (D) 1.6 J
Answer and explanation
Answer: (B). Find each kinetic energy first. K₀ = ½ × 0.80 × 3.0² = 3.6 J. K = ½ × 0.80 × 5.0² = 10.0 J. ΔK = 10.0 − 3.6 = 6.4 J. The direction does not matter, because K uses v².
- (A) uses ½m(Δv)² with Δv = 8.0 m/s. You cannot square a change in velocity.
- (C) gives the westward motion a negative kinetic energy. K is never negative.
- (D) uses ½m(Δspeed)² with 2.0 m/s. Subtract the energies, not the speeds.
If you missed this: read “Kinetic energy is a scalar” and Worked example 3 in the Topic 3.1 study guide.
Question 2 (multiple choice · 3.2)
A shopper pushes a loaded trolley 12 m along a level floor at constant velocity. Her push is 80 N, directed 25° below the horizontal. How much work do the resistive forces (friction) do on the trolley over the 12 m?
- (A) 0 J
- (B) −410 J
- (C) −870 J
- (D) −960 J
Answer and explanation
Answer: (C). The push does W = Fd cos θ = 80 × 12 × cos 25° = +870 J. The velocity is constant, so ΔK = 0 and W_net = 0. Gravity and the normal force are perpendicular to the motion, so they do no work. Friction must do −870 J.
- (A) mixes up the net work, which is zero, with the work done by one force.
- (B) uses sin 25°. That is the vertical part of the push, which does no work.
- (D) uses the whole 80 N and ignores the angle.
If you missed this: read “Net work and the work-energy theorem” and Worked example 1 in the Topic 3.2 study guide.
Question 3 (multiple choice · 3.2)
A 4.0 kg parcel is slid 5.0 m along a rough, level floor from P to Q, then slid back along the same line to P. Kinetic friction is 7.0 N each way. The parcel is then lifted 0.90 m onto a shelf. What is the total work done on the parcel by friction, and by gravity, for the whole trip?
- (A) friction 0 J; gravity −35 J
- (B) friction −70 J; gravity +35 J
- (C) friction −35 J; gravity −35 J
- (D) friction −70 J; gravity −35 J
Answer and explanation
Answer: (D). Friction is nonconservative. It opposes the motion on every metre of the 10 m path: −7.0 × 10 = −70 J. Gravity is conservative, so only the change in height counts: −mgh = −4.0 × 9.8 × 0.90 = −35 J. The sliding is level, so gravity does no work there.
- (A) uses the displacement for friction. The parcel ends where it started, but friction still acted all the way.
- (B) gets gravity’s sign wrong. The parcel rises while gravity points down.
- (C) counts only one leg of the slide.
If you missed this: read “Conservative and nonconservative forces” and “Where friction’s energy goes” in the Topic 3.2 study guide.
Question 4 (multiple choice · 3.3)
A satellite orbits a planet at 2R from the planet’s centre, where R is the planet’s radius. The planet–satellite system has gravitational potential energy −8.0 × 10⁹ J. The satellite moves to an orbit at 3R from the centre. What is the change in the system’s gravitational potential energy?
- (A) +2.7 × 10⁹ J
- (B) +4.4 × 10⁹ J
- (C) −2.7 × 10⁹ J
- (D) +5.3 × 10⁹ J
Answer and explanation
Answer: (A). U_g = −Gm₁m₂/r, so U_g ∝ 1/r. At 3R, U_g = −8.0 × 10⁹ × (2/3) = −5.3 × 10⁹ J. ΔU_g = −5.3 × 10⁹ − (−8.0 × 10⁹) = +2.7 × 10⁹ J. It is positive: moving apart raises the energy towards zero.
- (B) uses 1/r², which is how the force scales.
- (C) has the right size but the wrong sign. A less negative value is an increase.
- (D) is the size of the new value, not the change.
If you missed this: read “Gravitational potential energy” (the general form) in the Topic 3.3 study guide.
Question 5 (multiple choice · 3.4)
A student lowers a 5.0 kg box at constant speed from a shelf 1.2 m high to the floor. The system is box + Earth. Which statement is correct?
- (A) The system’s energy stays constant, because gravity is a conservative force.
- (B) 59 J of gravitational potential energy is converted into kinetic energy of the box.
- (C) The student’s hand does −59 J of work on the system, so the system’s energy falls by 59 J.
- (D) The student’s hand does +59 J of work on the system, because the hand pushes on the box.
Answer and explanation
Answer: (C). ΔU_g = −mgh = −5.0 × 9.8 × 1.2 = −59 J, and ΔK = 0. The hand is outside the system. It pushes up while the box moves down, so it does negative work: −59 J. The system’s energy changes by exactly that amount.
- (A) ignores the hand. A system’s energy is constant only if no outside force does work.
- (B) cannot be right: the speed is constant, so K does not change.
- (D) gets the sign wrong. The force and the displacement are opposite.
If you missed this: read “The energy rule” and “Choosing the system” in the Topic 3.4 study guide.
Question 6 (multiple choice · 3.5)
The kinetic energy of a 2.0 kg cart rises in a straight line on a graph of energy against time, from 0 at t = 0 to 36 J at t = 12 s. The track is level. Which statement is correct?
- (A) The power is a constant 3.0 W, and the net force on the cart decreases as it speeds up.
- (B) The power is a constant 3.0 W, and the net force on the cart is constant.
- (C) The power rises from 0 to 6.0 W, because P = Fv and the speed increases.
- (D) The power is 432 W.
Answer and explanation
Answer: (A). The slope of an energy–time graph is the power: 36 J ÷ 12 s = 3.0 W, the same at every moment. The speed rises (to 6.0 m/s at 12 s), so the net force F = P ÷ v must fall. At 12 s it is 3.0 ÷ 6.0 = 0.50 N.
- (B) A constant force with rising speed gives rising power, which would curve the graph upward.
- (C) assumes the force is constant. A straight energy–time line means constant power.
- (D) multiplies the energy by the time instead of dividing.
If you missed this: read “Reading graphs of energy and power” and Worked example 3 in the Topic 3.5 study guide.
Question 7 (short answer · 3.4)
A 0.50 kg block is released from rest at the top of a smooth curved ramp, 0.45 m above a level floor. On the floor it crosses a rough strip 1.5 m long (μ_k = 0.20), then hits a spring with k = 400 N/m on a smooth section. Everything else is frictionless.
(a) Find the block’s speed at the bottom of the ramp. (b) Find its speed as it leaves the rough strip. (c) Find the greatest compression of the spring. (d) The block bounces back. Where does it finally stop?
Worked answer
System: block + Earth + spring + floor, so friction is an internal conversion to thermal energy. Take U_g = 0 at the floor.
(a) mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 0.45) = 3.0 m/s (2.97 m/s). Start energy = 0.50 × 9.8 × 0.45 = 2.205 J.
(b) Friction = μ_k mg = 0.20 × 0.50 × 9.8 = 0.98 N. Energy dissipated on the strip = 0.98 × 1.5 = 1.47 J. K = 2.205 − 1.47 = 0.735 J, so v = √(2 × 0.735 ÷ 0.50) = 1.7 m/s.
(c) ½kΔx² = 0.735 J, so Δx = √(2 × 0.735 ÷ 400) = 0.061 m (6.1 cm).
(d) The block returns to the strip with 0.735 J. Each metre of strip removes 0.98 J, so it stops after 0.735 ÷ 0.98 = 0.75 m: the middle of the strip.
Check yourself: 1 point each for 3.0 m/s, the 1.47 J dissipated, 0.061 m, and 0.75 m with the reason.
If you missed this: read “When is mechanical energy constant?” and Worked example 3 in the Topic 3.4 study guide.
Question 8 (short answer · 3.3)
A 0.40 kg block hangs from a light vertical spring with k = 50 N/m. The spring starts at its relaxed length. A student holds the block and lowers it slowly until it hangs at rest on its own. The system is block + spring + Earth.
(a) Find how far the spring stretches. (b) Find the change in elastic potential energy and the change in gravitational potential energy. (c) The total potential energy of the system falls, but the block is at rest at the start and at the end. Explain where the energy went.
Worked answer
(a) At rest, kΔx = mg, so Δx = (0.40 × 9.8) ÷ 50 = 0.078 m (7.84 cm).
(b) ΔU_s = ½kΔx² = ½ × 50 × 0.0784² = +0.15 J. ΔU_g = −mgΔx = −3.92 × 0.0784 = −0.31 J. These are two separate pair terms: block–spring and block–Earth.
(c) The total change is 0.154 − 0.307 = −0.15 J, and ΔK = 0. The hand is outside the system. It pushes up while the block moves down, so it does −0.15 J of work: that energy leaves the system through the hand.
Check yourself: 1 point each for 0.078 m, +0.15 J with the square, −0.31 J with the sign, and the hand’s negative work as the reason.
If you missed this: read “Elastic potential energy of an ideal spring” and “Systems with more than two objects” (the block on a spring) in the Topic 3.3 study guide.
Your next step
Mark each question right or wrong. Count Questions 7 and 8 as right only if you earned every point.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 3.1 Translational kinetic energy | 1 | Topic 3.1 study guide |
| 3.2 Work | 2, 3 | Topic 3.2 study guide |
| 3.3 Potential energy | 4, 8 | Topic 3.3 study guide |
| 3.4 Conservation of energy | 5, 7 | Topic 3.4 study guide |
| 3.5 Power | 6 | Topic 3.5 study guide |
How to use your result
- Missed one question in a topic? Re-read the section named in that answer, then try the topic’s practice set.
- Missed two or more questions in a topic? Start again with that study guide, including its worked examples, before doing any mixed practice.
- Guessed? Treat the question as missed.
- Look for patterns. Errors in Questions 1 and 4 often come from signs: kinetic energy is never negative, but potential energy can be. Errors in Questions 2, 5 and 8 usually mean you did not ask which forces are outside the system.
- All correct? Move straight to the mixed unit review, which combines the topics in longer exam-style questions.