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

Answer:

The work done by the force is  5.76 J

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:

$K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J$

The final kinetic energy:

$K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J$

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