Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = t33, where x is in meter and t in seconds. The work done by the force in the first two seconds is:
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a
1.6 J
b
16 J
c
160 J
d
1600 J
answer is B.
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Detailed Solution
v = dxdt = ddt(t33) = t2When t = 0, then v = o, When t = 2, then v = 4 m/sWork done in first two seconds = change in KEW = 12m[(4)2-(0)2] = 122×16 = 16