The only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m/s in the positive x direction and some time later has a velocity of 6.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time?
By the work-kinetic energy theorem,
W=\Delta K=\frac{1}{2}m \nu_f^2-\frac{1}{2}m \nu_i^2=\frac{1}{2}(2.0 \,kg )\left((6.0 \,m / s )^2-(4.0 \,m / s )^2\right)=20 \,J.
We note that the directions of \vec{\nu}_f and \vec{\nu}_i play no role in the calculation.