A 3 kg object has initial velocity 6 i ^ - 2 j ^ m s - 1 . What will be the total work done (in joule) on the object if its velocity changes to ( 8 i ^ + 4 j ^ ) m s - 1 ?

Step-by-step solution

The net work done on the object is equal to the change in the kinetic energy of the object W n e t = K f - K i = Δ K Kinetic energy K = 1 2 m v 2 Velocity v 2 = v x 2 + v y 2 = ( 36 + 4 ) m 2 /s 2 v 2 = 40 m 2 /s 2 K f = 1 2 × 3 × 40 = 60 J Again, velocity v = 8 2 + 4 2 v = 64 + 16 = 80 m/s Kinetic energy K f = 1 2 3 × 80 m 2 /s 2 = 120 J From work energy theorem W n e t = K f - K i = 120 J - 60 J = 60 J