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A particle of mass 0.5kg travels in a st...

A particle of mass `0.5kg` travels in a straight line with velocity `v=ax^(3//2)` where `a=5m^(-1//2)s^-1`. What is the work done by the net force during its displacement from `x=0` to `x=2m`?

A

1.5J

B

50J

C

10J

D

100J

Text Solution

Verified by Experts

The correct Answer is:
B
Given,x`=ax^(3//2)`
`m=0.5kg,a=5m^(-1//2)s^(-1),"work done"(W)=?`
We know that
Acceleration `a_(0)=(dv)/(dt)=v(dv)/(dx)=ax^(3//2)(d)/(dx)"("ax^(3//2)")"`
`=ax^(3//2)xxaxx(3)/(2)xxx^(1//2)=(3)/(2)a^(2)x^(2)`
Now, Force=`ma_(0)=m(3)/(2)a^(2)x^(2)`
Work done= `int_(x=0)^(x=2)Fdx=int_(o)^(2)(3)/(2)ma^(2)x^(2)dx`
`=(3)/(2)ma^(2)xx(x^(3)//3)_(0)^(2)`
`(1)/(2),ma^(2)xx8=(1)/(2)xx(0.5)xx(25)xx8=50J`
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