An object of mass 10 kg is dropped from a tower of height 150 m. The body strikes the ground with velocity
40 ms–1. Calculate work done by air friction. [Take g = 10 m/s2]
Answers
Given : An object of mass 10 kg is dropped from a tower of height 150 m
The body strikes the ground with velocity 40 ms–1.
To Find : work done by air friction
Solution:
object of mass 10 kg is dropped from a tower of height 150 m
h = 150 m
v = 0
PE = mgh = 10 * 10 * 150 = 15000 J
KE = (1/2)mv² = 0 J
TE = PE + KE = 15000 + 0 = 15000 J
The body strikes the ground with velocity 40 ms–1.
v = 40 m/s
h = 0
PE = mgh = 0 J
KE = (1/2)mv² = (1/2)10(40)² = 8000J
TE = PE + KE = 0 + 8000 = 8000 J
work done by air friction = Change in Energy
= 8000 - 15000
= -7000 J
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