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
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