Question

The only force acting on a 2.5 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a velocity of 2.3 m/s in the positive x direction, and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 3.0 N force during this time?

Answer 1

Answer:44.58 J

Step-by-step explanation:

mass of block
`\left ( m\right )=2.5 kg`

Force magnitude=3 N

Initial velocity =
`2.3\hat{i} m/s`

Final velocity=
`6.4\hat{j} m/s`

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

=
`(1)/(2)* 2.5* 2.3^2=6.612 J`

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

=
`(1)/(2)* 2.5* 6.4^2=51.2 J`

Work Done =Final -Initial Kinetic energy=51.2-6.612=44.58 J