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=0.5kg, upsilon=ax^(3//2), a=5m^(-1//2)s^(-1), W=?`
Initial vel. At `x=0, upsilon=axx0=0, ` Final vel. At `x=2, upsilon_(2)=a2^(3//2)=5xx2^(3//2)`
Work done `=` increase in K.E. `=(1)/(2)m(upsilon_(2)^(2)-upsilon_(1)^(2))`
`W=(1)/(2)xx0.5[(5xx2^(3//2))^(2)-0]=50J`