An object of mass 4 kg is moving along x-axis such that its position x- varies with time ‘t’ as x2 = 16t3, with x in metre and t in sec. Work done as a function of time can be shown as
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a
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answer is C.
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Detailed Solution
given position x varying with time t x2 = 16t3 take square root both sides x = 4t32 differentiate x with respect to time t dxdt = 4×32 t1/2 dxdt = 6 t1/2---(1) dx=6 t1/2dt---(2) again differentiate eqn(1) with respect to time t ⇒d2xdt2 = 3t =acceleration---(3) work done dW=F.dx here force F=ma ⇒dW=ma .dx substitute equations (2),(3) and m=4kg in the above equation ⇒dW=43t6 t1/2dt ⇒dW=72 dt integrating ∫dW=∫72 dt ⇒W=72 t ⇒Work α time