A body of mass 1 kg thrown upwards with a velocity of 10 m/s comes to rest (momentarily) after moving up 4 m. The work done by air drag in this process is $(g = 10 m / s^{2})$

10 J

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- 10 J

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40 J

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50 J

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Solution

The correct option is B - 10 J
From work energy theorem $W_{g r} + W_{a i r d r a g} = Δ k \Rightarrow - m g h + W_{a i r d r a g} = 0 - \frac{1}{2} m u^{2} \Rightarrow W_{a i r d r a g} = m g h - \frac{1}{2} m u^{2} = (40 - 50) J = - 10 J$

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