Question

The only force acting on a 1.9 kg canister that is moving in an xy plane has a magnitude of 3.9 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 5.6 m/s in the positive y direction. How much work is done on the canister by the 3.9 N force during this time

Answer 1

Answer:

The work done on the canister is 15.34 J.

Explanation:

Given;

mass of canister, m = 1.9 kg

magnitude of force acting on x-y plane, F = 3.9 N

initial velocity of canister in positive x direction, $v_i$ = 3.9 m/s

final velocity of the canister in positive y direction, $v_j = 5.6 \ m/s$

The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.

ΔK.E = $W_{net}$

ΔK.E $= K.E_f -K.E_i$

The initial kinetic energy of the canister;

$K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J$

The final kinetic energy of the canister;

$K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J$

ΔK.E = 29.79 J - 14.45 J

ΔK.E = $W_{net}$ = 15.34 J

Therefore, the work done on the canister is 15.34 J.