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

The work done on the canister is 15.34 J.

Step-by-step 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 = (1)/(2) mv_i^2\\\\K.E_i = (1)/(2) m(√(v_i^2 +v_j^2 + v_z^2)\ )^2\\\\K.E_i = (1)/(2) *1.9(√(3.9^2 +0^2 + 0^2)\ )^2 = 14.45 \ J`

The final kinetic energy of the canister;

`K.E_f =(1)/(2) mv_j^2 \\\\K.E_f = (1)/(2) m(√(v_i^2 +v_j^2 + v_z^2)\ )^2\\\\K.E_f = (1)/(2) *1.9(√(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.