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`?

Explanation

Here, m=0.5 kg
`u=ax^(3//2), a=5m^(-1//2)s^(-1)`
Initial velocity at x=0, `v_(1)=axx0=0`
Final velocity at x=2, `v_(2)=a(2)^(3//3)= 5 xx (2)^(3//2)`
Work done = increase in K.E
`=1//2m(v_(2)^(2)-v_(1)^(2)) = 1//2 xx 0.5[(5 xx (2)^(3//3))^(2)-0]=50 J`