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Question

A force acts on a 30 g 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 meters and t is in seconds. The work done during the first 4 seconds is

A
5.28 J
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B
450 mJ
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C
490 mJ
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D
530 mJ
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Solution

The correct option is A 5.28 J
Velocity as a function of time is v=dxdt=38t+3t2
According to work energy theorem
W=12m(v24v20)
v4=38×4+3×42=19 ms1
v0=300=3 ms1

W=12×0.03×(19232)=5.28 J

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