Work, energy and power

WPH11Topic 1.43 questions

Work done, kinetic and gravitational potential energy, conservation of energy, power and efficiency.

Specification points

Be able to use the equation for work $\Delta W = F\Delta s$ , including calculations when the force is not along the line of motion
Be able to use the equation $E_k = \frac{1}{2}mv^2$

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21mv2

Be able to use the equation $\Delta E_{grav} = mg\Delta h$ for the difference in gravitational potential energy near the Earth's surface

Know, and understand how to apply, the principle of conservation of energy including use of work done, gravitational potential energy and kinetic energy

Be able to use the equations relating power, time and energy transferred or work done: $P = \frac{E}{t}$ and $P = \frac{W}{t}$

Be able to use the equations efficiency $= \frac{\text{useful energy output}}{\text{total energy input}}$ and efficiency $= \frac{\text{useful power output}}{\text{total power input}}$

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Questions

Multiple choice: 1
Calculation: 2
Total: 3

Sample questions

Original practice questions written in the style of the exam.

Question 12 marks
A car travels along a level road at a constant speed of $25\,\text{m s}^{-1}$ . The total resistive force on the car is $650\,\text{N}$ .

Calculate the useful power output of the car's engine.
Mark scheme
- At constant speed, driving force = resistive force $= 650\,\text{N}$ , and uses $P = Fv$ (1)
- $P = 650 \times 25 = 16\,000\,\text{W}$ ( $16\,\text{kW}$ ) (1)
Question 24 marks
A child on a sledge (total mass $15\,\text{kg}$ ) starts from rest and slides down a snowy slope. The bottom of the slope is $8.0\,\text{m}$ vertically below the start, and the length of the slope is $24\,\text{m}$ . At the bottom the speed is $9.5\,\text{m s}^{-1}$ .

m s−1

(a) Calculate the energy dissipated by resistive forces during the descent.

(b) Calculate the average resistive force acting on the sledge.

Mark scheme

Loss of GPE $= mgh = 15 \times 9.81 \times 8.0 = 1180\,\text{J}$ (1177 J) (1)
Gain in KE $= \tfrac{1}{2} \times 15 \times 9.5^2 = 677\,\text{J}$ (1)
Energy dissipated $= 1177 - 677 = 500\,\text{J}$ (1)
Average resistive force $= 500/24 = 21\,\text{N}$ (1)