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A particle moved from position vec r(1) ...

A particle moved from position `vec r_(1) = 3 hati + 2 hatj - 6 hatk `to position `vecr_(2) = 14 hati + 13 hatj +9 hatk` undre the action of a force `(4 hati + hatj +3 hatk)` newtons . Find the work done .

A

10j

B

100

C

0.01J

D

1J

Text Solution

Verified by Experts

The correct Answer is:
B
`therefore F=ma therefore a=F/m=(2t)/(1)`
But `a=(dv)/(dt)`
`therefore dv=adt=2tdt`
`therefore v=underset(0)overset(4)int 2tdt=[(2t^(2))/(2)]_(0)^(4)=16m//s`
Increase in K.E. of the body `=1/2mv^(2)-0`
`=1/2mv^(2) =1/2 xx 1 xx 16 xx 16=128J`
Work done by the force =Increase in K.E.=128J
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