- Calculate the work of a constant force with its sign, and the total work of several forces
- Use P = W/t and P = Fv, including a car's maximum speed at rated power
- Apply the work–energy theorem and conservation of mechanical energy with gravity and springs
- Find the heat produced by friction with Q = f · srel
Why does a car with the accelerator pressed to the floor still have a top speed? How can you find a skier's speed at the bottom of a slope without knowing its shape? Both answers come from one idea: forces transfer energy by doing work. The CSCA syllabus line «Work and energy, law of conservation of mechanical energy» is exactly this, and energy ideas return later in electricity, heat and modern physics. You will need Newton's second law from the lesson «Newton's laws of motion and their applications»; the next lesson, «Momentum, impulse and conservation of momentum», combines energy with momentum.
Work of a constant force
A force does work when its point of application moves and the force has a component along the displacement. Work is a scalar measured in joules: 1 J = 1 N · m.
- Wwork of the force, J
- Fmagnitude of the constant force, N
- smagnitude of the displacement, m
- αangle between the force and the displacement
Chinese textbooks write W = Fl cos α (l = displacement); the CSCA uses W for work and P for power, while some school textbooks use A and N. Only the component F cos α along the motion does work.
The sign of the work is decided by cos α. For 0 ≤ α < 90° the work is positive: the force helps the motion and gives the body energy. For α = 90° the work is zero. For 90° < α ≤ 180° the work is negative: the force resists the motion and takes energy away; we then say that the body does work against this force (Chinese 克服…做功).
| Angle α | Sign of W | Examples |
|---|---|---|
| 0 ≤ α < 90° | positive | a pulling force; gravity on a falling body |
| α = 90° | zero | the normal force on a level floor; the tension of a pendulum string |
| 90° < α ≤ 180° | negative | kinetic friction on a sliding body; gravity on a rising body |
1) A 20 N force pulls a sled 5 m along level snow; the rope makes 37° with the ground (cos 37° = 0.8). Find the work of the force.
2) A 2 kg bag is lifted 3 m vertically at constant speed. Find the work of the lifting force and of gravity (g = 10 m/s²).
3) A block slides 5 m while a kinetic friction force of 4 N acts on it. Find the work of friction.
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2) The lifting force equals mg = 20 N and points along the motion: W = 20 · 3 = +60 J. Gravity points against the motion (α = 180°): W = −20 · 3 = −60 J. The total work is 0, which matches the constant speed.
3) Friction is opposite to the motion: W = 4 · 5 · cos 180° = −20 J; the block loses 20 J of kinetic energy.
- Wtotaltotal work of all forces on the body (with signs)
- Fnetnet force (for constant forces)
Works are scalars: add them as signed numbers, not as vectors.
A 2 kg block is pulled 4 m along a horizontal floor by a horizontal force of 10 N. The coefficient of kinetic friction is 0.25 (g = 10 m/s²). What is the total work done on the block?
A) 40 J B) 20 J C) 60 J D) 0 J
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Total: 40 − 20 = 20 J → B.
A is only the work of the pulling force; C adds the friction work with a «+» sign; D comes from «gravity and the normal force cancel» while forgetting the pull and friction.
Power and a car at constant power
The rate of doing work, i.e. how fast energy is transferred. Unit: the watt, 1 W = 1 J/s; 1 kW = 1000 W.
- Ppower, W
- ttime taken, s
- vspeed of the body, m/s
- αangle between the force F and the velocity v
W/t gives the average power. F · v gives the instantaneous power with the instantaneous speed, or the average power with the average speed.
1) A motor lifts a 50 kg load at constant speed to a height of 12 m in 20 s. What is its power?
2) A 0.2 kg stone falls freely from rest. Find the power of gravity at t = 2 s and its average power over the first 2 s (g = 10 m/s²).
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2) At t = 2 s: v = gt = 20 m/s, P = mg · v = 2 · 20 = 40 W. The average speed over 2 s is (0 + 20)/2 = 10 m/s, so the average power is 2 · 10 = 20 W (check: in 2 s h = 20 m, W = 2 · 20 = 40 J, 40/2 = 20 W). In uniformly accelerated motion from rest the average power is half the final instantaneous power.
A car engine works at most at its rated power P. Since P = Fv, a higher speed means a smaller traction force F. With a resistive force f the acceleration a = (F − f)/m keeps falling while the car speeds up, and the speed stops growing when F has dropped to f. That is the car's maximum speed.
- Ftraction force of the engine, N
- fconstant resistive force, N
- vmaxmaximum speed on a level road, m/s
At vmax the acceleration is zero: the traction force exactly balances the resistance (Chinese textbooks call this topic 机车启动, «a vehicle starting off»).
In the second standard case (恒定加速度启动) the car first accelerates uniformly with a constant traction force F = f + ma. The power P = Fv reaches the rated power at v₁ = P/F: this is the end of the uniform acceleration, not the maximum speed. After that the car goes on at constant power with a decreasing acceleration up to vmax = P/f.
A 2000 kg car has a rated power of 80 kW; the resistive force is 4000 N. a) Find its maximum speed. b) Find its acceleration at 10 m/s at rated power. c) If it starts from rest with a constant acceleration of 0.5 m/s², how long does the uniform acceleration last?
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b) F = P/v = 80 000/10 = 8000 N; a = (8000 − 4000)/2000 = 2 m/s².
c) F = f + ma = 4000 + 2000 · 0.5 = 5000 N. Rated power is reached at v₁ = P/F = 80 000/5000 = 16 m/s, after t = v₁/a = 16/0.5 = 32 s. The car then keeps speeding up, more and more slowly, to 20 m/s.
Kinetic energy and the work–energy theorem
The energy of motion: Ek = mv²/2. It is a scalar, never negative, and depends on the speed but not on the direction of motion.
- Wtotaltotal work of all forces (gravity and friction included), J
- Ek₁, Ek₂initial and final kinetic energy
- v₁, v₂initial and final speed, m/s
The work–energy theorem (Chinese 动能定理, «kinetic energy theorem») also holds on curved paths and for changing forces, as long as you can find their work.
The theorem follows from Newton's second law and v² − v₀² = 2as: multiply F = ma by s and you get Fs = mv²/2 − mv₀²/2. Its big advantage: neither the time nor the acceleration is needed, and the path can be split into stages with different forces — just add the works of all stages.
- 1Choose the body and two states
Fix the start and the end; a body at rest has Ek = 0.
- 2List the forces
Write the work of each force with its sign: positive, negative or zero. Treat each stage with its own forces.
- 3Write the theorem
W₁ + W₂ + … = Ek₂ − Ek₁ and solve for the unknown.
- 4Check
Units and sign: distances and forces must come out positive; the mass often cancels.
1) A car moving at 20 m/s brakes with locked wheels; μ = 0.5, g = 10 m/s². Find the braking distance. What if the speed is 10 m/s?
2) A 10 g bullet flying at 300 m/s passes through a fixed board 5 cm thick and leaves it at 100 m/s. Find the average resistive force of the board.
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2) −F · d = mv₂²/2 − mv₁²/2 = 0.01 · (100² − 300²)/2 = −400 J ⇒ F = 400/0.05 = 8000 N.
A 2 kg block at rest on a horizontal floor is pushed 6 m by a horizontal force of 10 N, and then the force is removed. The coefficient of kinetic friction is 0.2 (g = 10 m/s²). How much farther does the block slide after the force is removed?
A) 15 m B) 3.6 m C) 9 m D) 18 m
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A forgets friction during the push (60/4 = 15 m); it is also the total distance (6 + 9). B divides the 36 J of kinetic energy by the 10 N push instead of by friction. D forgets the 2 in s = v²/(2μg) (v² = 36).
Potential energy and conservation of mechanical energy
Ep = mgh, where h is the height above a chosen reference plane (参考平面). Below the plane Ep is negative. Ep itself depends on the choice of the plane; its change does not.
- hheight above the reference plane, m
- WGwork of gravity when the body moves from height h₁ to h₂
- ΔEpchange of potential energy: Ep₂ − Ep₁
The work of gravity depends only on the start and end heights, never on the path: going down it is positive and Ep decreases; going up it is negative and Ep increases.
A 0.5 kg ball on a table 0.8 m high falls to the floor. Take as the reference plane a) the floor, b) the table top. Find the ball's Ep on the table and on the floor, and the work of gravity.
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b) On the table: 0; on the floor: 0.5 · 10 · (−0.8) = −4 J.
In both cases ΔEp = −4 J and WG = +4 J: the change does not depend on the reference plane.
The energy stored in a stretched or compressed spring: Ep = kx²/2, where x is the extension or compression measured from the natural length.
- kspring constant (stiffness), N/m
- xdeformation (extension or compression), m
Doubling the deformation stores four times the energy. The spring force kx itself belongs to the lesson «Forces and equilibrium».
A spring with k = 200 N/m is compressed first by 0.1 m, then by 0.2 m. How much energy is stored each time? Compressed by 0.1 m, the spring launches a 20 g ball along a smooth horizontal surface; find the ball's speed.
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All of the 1 J becomes kinetic energy: 0.02 · v²/2 = 1 ⇒ v² = 100 ⇒ v = 10 m/s.
The sum E = Ek + Ep is the mechanical energy (机械能). Gravity and spring forces only move energy between the kinetic and potential forms, so if no other force does work, this sum stays constant.
- Ek₁ + Ep₁mechanical energy in the first state
- Ek₂ + Ep₂mechanical energy in the second state
Condition: only gravity and elastic (spring) forces do work. Other forces may act if they do no work: the normal force of a smooth surface, the tension of a pendulum string.
1) A stone is thrown from a 15 m high cliff at 10 m/s in any direction. What is its speed when it hits the ground (no air resistance)?
2) A bob on a 0.8 m string is released from rest with the string horizontal. Find its speed at the lowest point. What if it is released with the string at 60° to the vertical?
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2) Height drop L = 0.8 m: v = √(2gL) = √16 = 4 m/s. At 60°: the drop is L(1 − cos 60°) = 0.4 m ⇒ v = √8 = 2√2 m/s ≈ 2.8 m/s. The tension is always perpendicular to the velocity, so it does no work.
A small ball is released from rest at the top of a smooth track 3.2 m high. What is its speed when it passes a point 1.4 m above the ground (g = 10 m/s²)?
A) 8 m/s B) 3√2 m/s C) 6 m/s D) 36 m/s
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A uses the whole 3.2 m (the speed at the ground); B forgets the 2 in 2gh (v² = 18); D forgets the square root.
Friction and heat: when mechanical energy is not conserved
When friction, air resistance or a pushing force does work, the mechanical energy changes. Chinese textbooks state this as the work–energy relation (功能关系): the work of all forces other than gravity and spring forces equals the change of mechanical energy. The mechanical energy lost to kinetic friction becomes heat.
- Wotherwork of the forces other than gravity and spring forces, J
- Emechanical energy: E = Ek + Ep
- Qheat produced by kinetic friction, J
- fkinetic friction force, N
- srelsliding distance of the two surfaces relative to each other (相对位移), m
The heat depends on the relative sliding distance, not on the distance measured from the ground. For a block sliding on a fixed floor, srel is simply the block's path.
1) A 1 kg box slides from rest down a rough ramp 5 m high and reaches the bottom at 8 m/s. How much heat is produced?
2) A conveyor belt moves at a constant 2 m/s. A 1 kg box is put on it at rest; μ = 0.2, g = 10 m/s². Find the heat produced while the box slips and the extra work done by the motor.
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2) a = μg = 2 m/s², so the box reaches 2 m/s after t = 1 s. In that time the box moves 1 m (average speed 1 m/s) and the belt 2 m, so srel = 1 m and Q = f · srel = 2 · 1 = 2 J. The box gains Ek = 1 · 2²/2 = 2 J. The motor must supply Ek + Q = 4 J (check: the 2 N friction on the belt × the belt's path of 2 m = 4 J).
Key terms and how CSCA asks about this
In the Chinese version of the test the same ideas appear under the names below. Even if you take the test in English, recognising them will help you at a Chinese university.
| Term | 中文 | Pinyin |
|---|---|---|
| work | 功 | gōng |
| positive / negative work | 正功 / 负功 | zhènggōng / fùgōng |
| power | 功率 | gōnglǜ |
| average / instantaneous power | 平均功率 / 瞬时功率 | píngjūn gōnglǜ / shùnshí gōnglǜ |
| rated power | 额定功率 | édìng gōnglǜ |
| kinetic energy | 动能 | dòngnéng |
| work–energy theorem | 动能定理 | dòngnéng dìnglǐ |
| gravitational potential energy | 重力势能 | zhònglì shìnéng |
| elastic potential energy | 弹性势能 | tánxìng shìnéng |
| reference plane | 参考平面 | cānkǎo píngmiàn |
| mechanical energy | 机械能 | jīxiènéng |
| law of conservation of mechanical energy | 机械能守恒定律 | jīxiènéng shǒuhéng dìnglǜ |
| relative displacement | 相对位移 | xiāngduì wèiyí |
| joule / watt | 焦耳 / 瓦特 | jiāo'ěr / wǎtè |
Questions on this topic usually follow one of these patterns; you have about 75 seconds for each:
- the work of a force at an angle, or the total work of several forces (watch the signs!);
- average vs instantaneous power; the power of gravity on a falling body; lifting at constant speed (P = mgv);
- a car at rated power: vmax = P/f, the acceleration at a given speed, the end of uniform acceleration;
- the work–energy theorem: stopping distance, average resistive force, «a push, then sliding»;
- conservation of mechanical energy: a smooth track, a pendulum, a spring launcher, the height where Ek = Ep; graphs of Ek and Ep against height (two straight lines crossing at H/2);
- heat from friction: lost mechanical energy, a conveyor belt, Q = f · srel.
Key points
- W = Fs cos α: positive for α < 90°, zero at 90°, negative for α > 90°; the total work is the signed sum.
- P = W/t is the average power, P = Fv gives the instantaneous power; at rated power a car's top speed is vmax = P/f.
- Work–energy theorem: Wtotal = Ek₂ − Ek₁, with the work of all forces.
- Ep = mgh (depends on the reference plane, its change does not) and Ep = kx²/2; the work of gravity WG = −ΔEp does not depend on the path.
- Mechanical energy is conserved when only gravity and spring forces do work; otherwise Wother = ΔE.
- Friction heat Q = f · srel uses the relative sliding distance.
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