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?

Respuesta :

Answer:15.66 J

Explanation:

mass of block [tex]\left ( m\right )=2.9 kg[/tex]

Force magnitude=7.5 N

Initial velocity =[tex]3.9\hat{i} m/s[/tex]

Final velocity=[tex]5.1 \hat{j} m/s[/tex]

Initial Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]

=[tex]\frac{1}{2}\times 2.9\times 3.9^2=22.05 J[/tex]

Final Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]

=[tex]\frac{1}{2}\times 2.9\times 5.1^2=37.714 J[/tex]

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