Question

The only force acting on a 2.9 kg canister that is moving in an xy plane has a magnitude of 7.5 N. The canister initially has a velocity of 3.9 m/s in the positive x direction, and some time later has a velocity of 5.1 m/s in the positive y direction. How much work is done on the canister by the 7.5 N force during this time?

Answer 1

Answer:15.66 J

Explanation:

mass of block $\left ( m\right )=2.9 kg$

Force magnitude=7.5 N

Initial velocity =$3.9\hat{i} m/s$

Final velocity=$5.1 \hat{j} m/s$

Initial Kinetic Energy=$\frac{1}{2}mv^2$

=$\frac{1}{2}\times 2.9\times 3.9^2=22.05 J$

Final Kinetic Energy=$\frac{1}{2}mv^2$

=$\frac{1}{2}\times 2.9\times 5.1^2=37.714 J$

Work Done =Final -Initial Kinetic energy=37.714-22.056=15.66 J