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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» Answer: Answer Given, M=1800kg u=10ms −1
s=50m F=? We know that 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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