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

= 0.5 ` a = 5 m^(-1//2) s^(-1)`
`v=ax^(3//2)`
At x = 0 , `v_(1) = a. 0 = 0 `
At x = 2 , `v_(2) = a.(2)^(3//2) = 5xx(2)^(3//2)`
Work done = Change in kinetic energy
`1/2mv_(2)^(2) - 1/2mv_(1)^(2)`
`= 1/2 m(v_(2)^(2) - v_(1)^(2))`
`= 1/2 xx 0.5 xx 5 xx 5 xx 2 xx 2xx 2 `
` = 50 J `