Question

# 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 $$x=3t-4t^2+t^3$$, where x is in metres and t is in seconds. The work done during the first $$4$$ second is?

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

Solution

## The correct option is A $$5.28$$J$$M=20gm=0.020kg$$$$x=3t-4t²+t³m$$$$v=\frac { dx }{ dt } =3-8t+3t²\frac { m }{ s }$$$$a=\frac { dv }{ dt } =-8+6t\frac { m }{ { s }^{ 2 } }$$Force$$=m\ast a=0.120t-0.160N$$Workdone$$=dW=Fdx=F\ast \frac { dx }{ dt } \ast dt=(0.120t-0.160)\ast (3-8t+3t²)dt=0.360t³-1.440t²+1.640t-0.480$$$$W=0.090t⁴-0.490t³+0.82t²-0.480t=2.88J$$Physics

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