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=3t4t2+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=38t+3t2

v0=3m/s  and  v4=19m​​/s

W=12m(v42v02)  (According to work energy theorem)

=12×0.03×(19232)=5.28J

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