A force acts on a 30 gm 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 during the first 4 seconds is
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a
5.28 J
b
450 mJ
c
490 mJ
d
530 mJ
answer is A.
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Detailed Solution
v=dxdt=3−8t+3t2∴v0=3m/s and v4=19m/sW=12mv42−v02 (According to work energy theorem)=12×0.03×192−32=5.28J
A force acts on a 30 gm 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 during the first 4 seconds is