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Questions on Newton's Laws

Q1.

The diagram shows an electric two-wheeled vehicle and driver.

(a) The vehicle accelerates horizontally from rest to 27.8 m s–1 in a time of 4.6 s.

The mass of the vehicle is 360 kg and the rider has a mass of 82 kg.

(i) Calculate the average acceleration during the 4.6 s time interval. Give your answer to an appropriate number of significant figures.

average acceleration = (v-u)/t

a = 27.8/4.6 = 6.04 m s-2

a = 6.0 m s-2

(to the correct number of significant figures. Within the question there is information to 3sf (final velocity) but most of the info is only to 2sf - always use the minimum number of sf in the information to determine how many you should quote the final answer to!)

[2 marks]

(ii) Calculate the average horizontal resultant force on the vehicle while it is accelerating.

F = ma

m = 360 + 82 = 442 kg

a = 6.04 m s-2

F = 442 x 6.04

F = 2700 N

[2 marks]

(b) State and explain how the horizontal forward force on the vehicle has to change for constant acceleration to be maintained from 0 to 27.8 m s–1.

For constant acceleration there must be a constant net force acting on the system. The driving force is provided by the engine of the car but the counter forces are due to fair resistance (or drag). This depends on the speed of the vehicle. As the vehicle gets faster the drag increases. This would result in a net decrease of resultant forward force, so to maintain the acceleration the forward driving force would have to increase producing a constant net driving force. (3 max)

[3 marks]

(c) The electric motors drive both wheels of the vehicle. Add labelled force arrows to diagram of the vehicle to show the horizontal forces acting on the vehicle when it is moving at a constant speed.

The drag arrow can be anywhere on the car, but the driving force ones must come from BOTH wheels and be at the point of contact between the wheel and road or at the axle.

Arrows that were not horizontal got no mark.

Arrows labelled friction were ignored. The examiners wanted drag force or air resistance - 'wind resistance', 'resistance' or 'friction force' got no marks!

One mark for drag - one for driving force

[2 marks]

 

(d) The vehicle now accelerates to a constant speed of 55 m s–1. The useful power output of the motors is 22 kW at this speed.

Calculate the horizontal resistive force acting on the vehicle.

P = Fv

F = P/v

F = 22 x 10-3/55

F = 400 N

[2 marks]

(11 marks Total)