The only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 6.0 N. The canister initially has a velocity of 2.2 m/s in the positive x direction, and some time later has a velocity of 7.9 m/s in the positive y direction. How much work is done on the canister by the 6.0 N force during this time

W_net = 57.57J

Explanation:

We are given;

Mass of canister; m = 2kg

Initial velocity of the canister; v_i = 2.2 m/s

Final velocity of the canister; v_f = 7.9 m/s

From work energy theorem, the net work done is equal to the change in kinetic energy. Thus;

W_net = ΔKE

W_net = KE_f - KE_i

W_net = (1/2)mv_f² - (1/2)mv_i²

W_net = (1/2)m(v_f² - v_f²)

Plugging in the relevant values to give;

W_net = (1/2)(2)(7.9² - 2.2²)

W_net = 57.57J

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the answer is 3.1 his answer is wrong mines is right