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A force acts on a 3.0 gm particle in suc...

A force acts on a 3.0 gm particle in such a way that the position of the particle as a function of time is given by ` x=3t-4t^(2)+t^(3)`, where xx is in metres and t is in seconds. The work done during the first 4 seconds is

A

450 m

B

490 mJ

C

570 mJ

D

528 mJ

Text Solution

Verified by Experts

The correct Answer is:
D
`therefore x=t^(3)/3 therefore " Velocity "=(dx)/(dt)=(3t^(2))/(3)=t^(2)` ltbgt `therefore" For "t=2s, v=2^(2)=4m//s`
Work done `=F xx s`
But as both F and s are not given, we use the work energy theorem.
Work done=Change in K.E `=1/2 mv^(2) ("as the block starts from rest")`
`=1/2 xx 2 xx 4^(2)=16J`
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