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

Options

A. `1.5 J`
B. 50 J
C. 10 J
D. 100 J

Answer

B

Explanation

Given, `v = ax^(3//2)`
We know that
Acceleration
`a_(0)=(dv)/(dt) = v(dv)/(dx) = ax^(3//2) (d)/(dx) (ax^(3//2))`
=`ax^(3//2) xxa xx(3)/(2) xx x^(1//2) = (3)/(2) a^(2) x^(2)`
Now, Force `= ma_(0) = m (3)/(2) a^(2) x^(2)`
Work done `= underset(x=0) overset(x=2) int Fdx = underset(0) overset(2) int(3)/(2) ma^(2)x^(2) dx`
=`(3)/(2) ma^(2) xx(x^(3)//3)_(0)^(2)`
=`(1)/(2) ma^(2) xx 8=(1)/(2) xx(0.5)xx(25)xx8=50 J`.