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5 May, 17:36

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

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  1. 5 May, 17:52
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    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
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