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Question

A block of mass 3 kg is released from the rest at top of the inclined plane. If length and height of the inclined plane is 6 m and 4 m respectively. Calculate its speed at the bottom of inclined plane. Assume the force of friction 16 N on the block is a constant.

A
2 m/s
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B
3 m/s
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C
4 m/s
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D
5 m/s
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E
6 m/s
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Solution

The correct option is A 4 m/s
Acceleration of the block along the inclined surface=a=gsinθfm
=10m/s2×4616N3kg
=43m/s2
Distance traveled along the surface=s=6m
Thus speed attained by the block=2as=4m/s

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