Work & Power — Practice · Grade 12 Science

Chapter: Work, Energy & Momentum

What this practice covers

By the end of Work & Power you should be able to:

9 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: work, joule (J), work-energy theorem, power, watt (W).

Problem 1

A 30 N force pushes a box 3.0 m in the direction of the force. How much work is done?
Answer and explanation

90 J

W = Fd = 30 × 3.0 = 90 J.

Problem 2

A crane lifts a 150 kg steel beam 8.0 m upward at constant speed. How much work does it do against gravity? (g = 9.8 m/s²)
Answer and explanation

11760 J

The lifting force equals the weight, mg = 150 × 9.8 = 1,470 N. W = Fd = 1,470 × 8.0 = 11,760 J.

Problem 3

Two cranes lift identical loads to the same rooftop, but Crane A takes 4 s while Crane B takes 8 s. Which statement is correct?
  1. Crane B does more work on the load
  2. The cranes have equal power
  3. Crane A has greater power
  4. Crane A does more work on the load
Answer and explanation

Crane A has greater power

Both cranes do the same work (same load, same height). Since P = W/t, doing that work in less time means Crane A delivers greater power.

Problem 4

An elevator motor does 18,000 J of work in 6.0 s. Find its power output.
Answer and explanation

3000 W

P = W/t = 18,000 ÷ 6.0 = 3,000 W.

Problem 5

The net force on a 3.0 kg cart, initially at rest, does 150 J of work on it. Using the work-energy theorem, find the cart's final speed.
Answer and explanation

10 m/s

W_net = ΔKE, so 150 = ½ × 3.0 × v² = 1.5v². Then v² = 100 and v = 10 m/s.

Problem 6

A device does 900 J work in 30 s. Find average power. Show the reasoning, not just the conclusion.
Hints

State which force's work is calculated.

Answer and explanation

900/30=30 W.

Problem 7

When can work by a force be negative?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

When its component along displacement opposes the displacement, such as kinetic friction on a sliding object.

Problem 8

Power measures the rate of energy transferPower measures the rate of energy transferTransferred work (J)Machine A: 3 s600Machine B: 6 s600PA = 600/3 = 200 WPB = 600/6 = 100 WEqual work can be transferred at different rates.
Illustrative equal-work comparison; durations are supplied, not inferred from bar widths.
Interpret the supplied visual. How much work does a steady 100 W device transfer in 12 s?
Answer and explanation

W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.

Power is energy transferred per time: average P=W/Δt for this process. A watt is one joule per second. Two machines can perform the same work while having different power because one takes less time. A power rating alone does not give total energy without duration and actual operation. Distinguish average power over a stated interval from brief peak power and keep seconds consistent when computing watts.

W=PΔt=1200 J, assuming the stated steady output applies for all 12 s.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

A motor raises a 20 kg load by 3 m in 6 s at 70% efficiency. Find useful work, output power and input energy.
Answer and explanation

Useful work=mgh=588 J. Output power=98 W. Input energy=588/0.70=840 J; other transfers/storage total 252 J under the stated boundary.

Useful work=mgh=588 J.

Output power=98 W.

Input energy=588/0.70=840 J; other transfers/storage total 252 J under the stated boundary.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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