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A block of mass 100kg is pulled up on an inclined plane of length 10m and 3m high by a constant force of 400N. If the work done against friction is 200J, the velocity of the block at the top of the inclined plane is (g = $10 m s^{- 2}$ )

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$8 m s^{- 1}$

$1 m s^{- 1}$

$4 m s^{- 1}$

$2 m s^{- 1}$

answer is B.

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Detailed Solution

$F . S = m g h + f . S + \frac{1}{2} m v^{2}$

$400 \times 10 = 100 \times 10 \times 3 + 200 + \frac{1}{2} \times 100 \times v^{2}$

$\Rightarrow V = 4 \frac{m}{s}$

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