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

(b) Given, `v=ax^(3//2)`
m=0.5 kg, a=5 `m^(-1//2)s^(-1)`,work done (W)=?
We know that
Accelaration `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_(0)^(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`