A body of mass 2 kg is falling freely through a height of 20 m. Velocity just before hitting ground is 15 m/s, then work done by air drag is

175 J

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-175 J

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575 J

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-225 J

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Solution

The correct option is B -175 J
By $W_{E T} \Rightarrow W_{G} + W_{f} = Δ K . E$
$m g H + W_{f} = \frac{1}{2} m v^{2} - 0$ and $W_{f} = \frac{m v^{2}}{2} - m g H$
$W_{f} = \frac{2 \times 15^{2}}{2} - 2 \times 10 \times 20 = - 175 J$

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A body of mass 2 kg is falling freely through a height of 20 m. Velocity just before hitting ground is 15 m/s, then work done by air drag is

The correct option is B -175 JBy WET⇒WG+Wf=ΔK.E mgH+Wf=12mv2−0 and Wf=mv22−mgH Wf=2×1522−2×10×20=−175 J

The correct option is B -175 J
By $W_{E T} \Rightarrow W_{G} + W_{f} = Δ K . E$
$m g H + W_{f} = \frac{1}{2} m v^{2} - 0$ and $W_{f} = \frac{m v^{2}}{2} - m g H$
$W_{f} = \frac{2 \times 15^{2}}{2} - 2 \times 10 \times 20 = - 175 J$