Certain force acting on a 20 kg mass changes its velocity from `5m//s to 2m//s`. calculate the work done by the force.

Explanation

Here, mass of the body, `m = 20 kg`
initial velocity `u = 5m//s`
final velocity, `upsilon = 2 m//s`
Initial kinetic energy of the body,` E_(ki) = (1)/(2) (20kg)(5m//s)^2 = 250J`
Final kinetic energy of the body, `E_(kf) = (1)/(2)m upsilon^2 = (1)/(2) (20kg) (2m//s)^2 = 40J`
Work done by the force = change in kinetic energy
`=E_(ki) - E_(kf) = 250 J - 40J = 210 J`
According to work energy theorem,
`W = (1)/(2) mu^2 =- (1)/(2) m upsilon^2 o= (1)/(2) m (u^2 - upsilon^2) = (1)/(2) = (20kg) [ (5 m//s)^2 - (2m//s)^2)]`
`=(1)/(2) (20kg) (21 m//s)^2] = 210 J`