The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 2.6 N. The canister initially has a velocity of 2.3 m/s in the positive x direction, and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 2.6 N force during this time?

55.29 J

Explanation:

mass of canister, m = 3.1 kg

magnitude of force, F = 2.6 N

initial velocity, u = 2.3 m/s

final velocity, v = 6.4 m/s

Let W is the work done.

According to the work energy theorem, the work done by all forces is equal to he change in kinetic energy.

here F is the only force acting on the canister.

initial kinetic energy, k = 0.5 x m x u²

k = 0.5 x 3.1 x 2.3 x 2.3 = 8.1995 J

final kinetic energy, k' = 0.5 x m x v²

k' = 0.5 x 3.1 x 6.4 x 6.4 = 63.488 J

So, the change in  kinetic energy,

Δk = k' - k

Δk = 63.488 - 8.1995 = 55.29 J

Thus, the work done by the force is 55.29 J.

for this one it would be d bump/forearm pass