Question 1Multiple choice · 1 mark
A cyclist produces 300 W while travelling at 10 m s−1. What is the total resistive force?
- A 3.0 N
- B 30 N
- C 300 N
- D 3000 N
Show answer
Answer: B
Free Work, energy & power questions for AS Level Physics (9702), part of Mechanics & materials. Each question has the answer and the marking points an examiner would look for.
Generate a full Work, Energy & Power worksheet →All Mechanics & materials questionsA cyclist produces 300 W while travelling at 10 m s−1. What is the total resistive force?
Answer: B
Which term is described by this statement?
“The product of force and displacement in the direction of the force.”
Answer: D
A ball of mass 0.8 kg is dropped from a height of 1.5 m. It rebounds to a height of 1.05 m.
Calculate the speed just before impact.
Calculate the energy lost in the bounce.
The ball is in contact with the ground for 0.050 s. Calculate the average force exerted by the ground.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 5.42 m s−1 mgh = ½mv2 C1 5.42 m s−1 A1 | 2 |
| (b) | 3.53 J mg(h − h′) C1 3.53 J A1 | 2 |
| (c) | 167 N change in momentum m(v1 + v2) C1 ÷ 0.050 s C1 add weight: 167 N A1 | 3 |
A car of mass 1500 kg moves up a slope inclined at 5° to the horizontal at a constant speed of 15 m s−1. The total resistive force is 600 N.
Calculate the component of the car’s weight down the slope.
Calculate the useful output power of the engine.
Calculate the rate of increase of gravitational potential energy of the car.
The engine is 30% efficient. Calculate the input power.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1280 N 1500 × 9.81 × sin5° = 1280 N B1 | 1 |
| (b) | 28200 W driving force = 1280 + 600 C1 = 1880 N C1 P = Fv = 28200 W A1 | 3 |
| (c) | 19200 W mg × vertical speed = 1500 × 9.81 × 15sin5° C1 19200 W A1 | 2 |
| (d) | 94100 W 94100 W B1 | 1 |
Define work done and power.
Define work done by a force.
Derive the expression for gravitational potential energy, Ep = mgh, from the definition of work.
Define efficiency.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | force × displacement in direction of force product of force and displacement in the direction of the force B1 | 1 |
| (b) | W = Fs with F = mg and s = h work done against weight = force × distance = mg × h B1 this equals the gain in GPE B1 | 2 |
| (c) | useful output ÷ total input useful energy (power) output ÷ total energy (power) input B1 | 1 |
A car of mass 1500 kg travels at a constant speed of 30 m s−1 along a straight horizontal road. The useful output power of the engine is 30 kW.
Define power.
Calculate the total resistive force acting on the car.
The car now climbs a slope inclined at 5° to the horizontal at the same speed, with the same resistive force. Calculate the output power now required.
Calculate the gain in gravitational potential energy per second on the slope.
The efficiency of the engine is 30%. Calculate the input power on the level road.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | rate of doing work work done per unit time B1 | 1 |
| (b) | 1000 N P = Fv C1 F = 1000 N A1 | 2 |
| (c) | 68.5 kW component of weight along slope = 1500 × 9.81 × sin 5° = 1280 N C1 total force = 2280 N C1 P = 68.5 kW A1 | 3 |
| (d) | 38500 W height gained per second = 30 sin 5° C1 38500 J s−1 A1 | 2 |
| (e) | 100 kW 30/0.30 = 100 kW B1 | 1 |
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