Under the action of a force a 4kg body moves such that its position ‘x ’ as a function q first x=t33 when ‘x ’ in m/s and ‘t ’ in seconds. The work done by force first ‘3’ seconds
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a
162 J
b
9 J
c
16 J
d
4 J
answer is A.
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Detailed Solution
Given mass of body m=4 kg Displacement x=t33 Velocity v=dxdt=ddtt33=t2=32=9 m/sec Work done W=ΔKE=12mv2−12mu2 (∵u=0) W=12×4×92−0=162 J