A car travels at a constant speed of 25 m s-1 up a slope. The wheels driven by the engine exert a forward force of 3000 N.

Question 21

A car travels at a constant speed of 25 m s^-1 up a slope. The wheels driven by the engine exert a forward force of 3000 N. There is a drag force due to air resistance and friction of 2100 N. The weight of the car has a component down the slope of 900 N.

What is the rate at which thermal energy is dissipated?

A zero              B 2.3 × 10⁴ W           C 5.3 × 10⁴ W           D 7.5 × 10⁴ W

Reference: Past Exam Paper – November 2016 Paper 12 Q20

Solution:

Answer: C.

When asked for ‘the rate at which thermal energy is dissipated’, we are actually being asked for the quantity representing ‘rate of energy’ or ‘energy / time’ which is the power.

So, we are being asked to find the ‘power’ dissipated.

Different forces act on the car but we need to identify which of these forces causes thermal energy to be dissipated.

Air resistance and friction both cause a dissipation of thermal energy, so we need to consider the 2100 N force only.

For a body moving at constant speed,

Power = Force × Speed

P = Fv

Power dissipated = 2100 × 25

Power dissipated = 5.3 × 10⁴ W                     [C is correct]

It is important to understand the difference between force and energy. It can be noticed that the forces acting on the car are balanced (forward force = 3000 N and backward force = 2100 + 900 = 3000 N) and that the resultant force is zero. Many would use the resultant force as zero to obtain answer A.

However, only one of these forces cause a dissipation of thermal energy.

But, consider the energy transformation that occurs. The net change during the motion of the car is from chemical energy in the fuel to gravitational potential energy of the car and thermal energy.