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

Explanation:

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Answer 2

Answer:

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

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 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