A particle of mass m kg moves along the X-axis with its velocity varying with the distance travelled as `v= kx^(beta)`, where k is a positive constant. The total work done by all the forces during displacement of the particle for x = 0 to x=d is close to

Options

A. `(mk^(2))/(2)`
B. `(mk^(2))/(2) d^(2 beta)`
C. `(mk^(2))/(2 beta)`
D. `(mk^(2) d)/(2 beta)`

Answer

B

Explanation

Velocity , `v= kx^(beta)`
By work-energy theorem,
Work done = Change in KE
= Final KE - Initial KE
`= (1)/(2) mv_(f)^(2) - (1)/(2) mv_(i)^(2)`
`=(1)/(2) m(kd^(beta))_(2) - (1)/(2) m(k xx 0)^(2) = (mk^(2))/(2). D^(2beta)`