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`?

Options

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

Answer

B

Explanation

Here,`m=0.5kg, upsilon=ax^(3//2), `where`a=5m^(-1//2), W=?`
Acceleration, `A=(dupsilon)/(dt)=(d)/(dt)(ax^(3//2))=axx(3)/(2)x^(1//2)(dx)/(dt)=(3a)/(2)x^(1//2)(ax^(3//2))`
`A=(3a^(2))/(2)x^(2)`
`F=mA=0.5xx(3a^(2))/(2)x^(2)=(3a^(2))/(4)x^(2)`
`W=int_(0)^(2)F.dx=int_(0)^(2)(3a^(2))/(4)x^(2)dx=(3a^(2))/(4)[(x^(3))/(3)]_(0)^(2)=(3)/(4xx3)(5)^(2)[2^(3)-0]=50J.`