A car of mass 1500kg is moving with a speed of 10m/s brought to rest after applying breaks. How much work is done by breaks?
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Answer
Given,
M=1800kg
u=10ms
−1
s=50m
F=?
We know that
V
2
−u
2
=2as
0−(10)
2
=2a×50m
a=
100
−100
=−1ms
−2
F=ma
Therefore, deceleration=−1m/s
2
Deceleration=−(acceleration)
F=1800×1ms
−2
=1800N
Force will be acting in the opposite direction to a displacement of the car
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