Question

The only force acting on a 3.2 kg canister that is moving in an xy plane has a magnitude of 6.7 N. The canister initially has a velocity of 3.3 m/s in the positive x direction, and some time later has a velocity of 6.9 m/s in the positive y direction. How much work is done on the canister by the 6.7 N force during this time

Answer 1

The work done by the force is 5.76 J

Step-by-step explanation:

Given;

mass of canister , m = 3.2 kg

magnitude of force, f = 6.7 N

initial velocity of the canister on x-axis,
`v_i`= 3.3i m/s

final velocity of the canister on y- axis,
`v_f` = 6.9j m/s

The work done on the canister = change in the kinetic energy of the canister

`W = K.E_f - K.E_i`

where;

K.Ei is the initial kinetic energy

K.Ef is the final kinetic energy

The initial kinetic energy:

`K.E_i = (1)/(2) *m√(i^2 +j^2+z^2)\\\\K.E_i = (1)/(2) *3.2√(3.3^2 +0^2+0^2)\\\\K.E_i = 5.28 \ J`

The final kinetic energy:

`K.E_f = (1)/(2) *m√(i^2 +j^2+z^2)\\\\K.E_f = (1)/(2) *3.2√(0^2 +6.9^2+0^2)\\\\K.E_f = 11.04 \ J\\`

W = 11.04 - 5.28

W = 5.76 J

Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J