Question

The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 3.6 m/s in the positive x direction, and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time

Answer 1

Step-by-step explanation:

Below is an attachment containing the solution.

Answer 2

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Step-by-step explanation:

Let the initial kinetic energy of the canister be

KE₁ =
`(1)/(2) mv_1^(2)` =
`(1)/(2) *3*3.6^(2)` = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ =
`(1)/(2) mv_2^(2)` =
`(1)/(2) *3*7.0^(2)` = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J