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Q.

A block of mass 2 kg rests on ground. An external agent raises the block to a height of 5m above ground where its speed is V. It work done by the external agent is 200 J, value of V is

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a

10 m/s

b

5 m/s

c

4 m/s

d

20 m/s

answer is A.

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Detailed Solution

Work done by the external agent = Change in kinetic energy + change in P.E.∴ 200=12mV2+mgh⇒ 200=12×2×V2+2×10×5⇒V=10 m/s

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