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A body of mass 4kg is freely, released from certain height where air resistive force of 8N is acting on the body. Considering its motion for first 2 seconds, work done by air resistive force is _____ $(g = 10 {ms}^{- 2})$

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160 J

– 160 J

128 J

– 128 J

answer is D.

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Detailed Solution

$F_{Net} = Ma$

$40 - 8 = 4 a \Rightarrow a = 8 m / s^{2}$

$S = ut + \frac{1}{2} {at}^{2}$

$d = 0 + \frac{1}{2} 8 {(2)}^{2} = 16 m ↓$

Work done by $F_{r} = F_{r} d \cos θ$

$= 8 (16) \cos 180 °$

$= (128) (- 1) = - 128 J$

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