A cyclist is travelling at a constant speed up a hill. The frictional force resisting the cyclist’s motion is 8.0 N.

Question 28

A cyclist is travelling at a constant speed up a hill. The frictional force resisting the cyclist’s motion is 8.0 N.

The cyclist uses 450 J of energy to travel a distance of 20 m.

What is the increase in the gravitational potential energy of the cyclist?

A 160 J                        B 290 J                        C 440 J                       D 610 J

Reference: Past Exam Paper – June 2017 Paper 12 Q15

Solution:

Answer: B.

Since the cyclist is moving at constant speed, he is NOT gaining kinetic energy.

From conservation of energy,

Work done by cyclist = Gain in GPE + Work done against friction

450 = Gain in GPE + (8.0×20)

450 = Gain in GPE + 160

 
Gain in GPE = 450 – 160 = 290 J