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

Step-by-step 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=
`(1)/(2)mv^2`

=
`(1)/(2)* 2.9* 3.9^2=22.05 J`

Final Kinetic Energy=
`(1)/(2)mv^2`

=
`(1)/(2)* 2.9* 5.1^2=37.714 J`

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