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Q.

A force acts on a 30g particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3 , where x is in metres and t is in seconds. The work done (in J) during the first 4 seconds  is

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answer is 5.28.

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Detailed Solution

v=dxdt=3−8t+3t2∴v0=3 m/s  and  v4=19 m​​/​sW=12m(v42−v02)  (According to work energy theorem)=12×0.03×(192−32)=5.28 J
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