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, [tex]v_i[/tex] = 3.9 m/s
final velocity of the canister in positive y direction, [tex]v_j = 5.6 \ m/s[/tex]
The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.
ΔK.E = [tex]W_{net}[/tex]
ΔK.E [tex]= K.E_f -K.E_i[/tex]
The initial kinetic energy of the canister;
[tex]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[/tex]
The final kinetic energy of the canister;
[tex]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[/tex]
ΔK.E = 29.79 J - 14.45 J
ΔK.E = [tex]W_{net}[/tex] = 15.34 J
Therefore, the work done on the canister is 15.34 J.
