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Question

A force acts on a 3 gm 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 second is:

A
384mJ
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B
168mJ
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C
528mJ
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D
541mJ
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Solution

The correct option is C 528mJ
Mass of the particle,
m=0.003kg
x=3t4t2+t3
a=8+6t
Thus force acting on the particle is ma
Thus the work done on the particle=Fdx
=40m(8+6t)(38t+3t2)dt
=528mJ -m represents mass here.


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