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

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, = 3.9 m/s

final velocity of the canister in positive y direction,

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

ΔK.E =

ΔK.E

The initial kinetic energy of the canister;

The final kinetic energy of the canister;

ΔK.E = 29.79 J - 14.45 J

ΔK.E = = 15.34 J

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

f=ma newton

f=100x2.5=250n