Question 39
An object of mass 0.30
kg is thrown vertically upwards from the ground with an initial velocity of 8.0
m s-1. The object reaches a maximum height of 1.9 m.
How much work is done
against air resistance as the object rises to its maximum height?
A 4.0 J B 5.6
J C 9.6
J D 15
J
Reference: Past Exam Paper – June 2016 Paper 13 Q17
Solution:
Answer: A.
The object is thrown with
an initial velocity of 8.0 m s-1. So, it has some initial kinetic
energy.
At the object moves up,
its kinetic energy is converted into gravitational potential energy. Some of
the kinetic energy is expended as work done against air resistance.
KE at bottom = GPE at max
height + Work done against air resistance
KE at bottom = ½ mv2
= ½ × 0.30 × 82
= 9.6 J
GPE at top = mgh = 0.30 ×
9.8 × 1.9 = 5.586 J
KE at bottom = GPE at max
height + Work done against air resistance
9.6 = 5.586 + Work done
against air resistance
Work done against air
resistance = 9.6 – 5.586 = 4.014 J
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation