A body of mass 600 g travels in a straight line with velocity `v=ax^((3)/(2))`. Where `a=3m^((1)/(2))s^(-1)`. What is the workdone by the net force during its displacement from x = 0 to x = 3 m ?

Explanation

Mass of a body `(m)=(600)/(1000)kg=0.6kg`
Velocity `(v)=ax^((3)/(2))` where `a=3m^(1//2)s^(-1)`
We know that work done by net force acting on a body is equal to its change in K.E. If velocity of body corresponding to x = 0 and x = 3 m be u and v respectively than
workdone `W=(1)/(2)mv^(2)-(1)/(2)mv^(2)=(1)/(2)m(v^(2)-u^(2))`
Initial velocity `u=a(0)^(3//2)=0` and `v=a.(3)^(3//2)`
`W=(1)/(2)xx0.6[{a(3)^(3//2)}^(2)-0]`
`=(1)/(2)xx0.6xx a^(2)xx(3^(3//2))^(2)=(1)/(2)xx0.6xx3^(2)xx(3)^(3)`
Workdone `W=(1)/(2)xx0.6xx9xx27=73 J`