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2 August, 20:29

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?

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  1. 2 August, 22:50
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    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.
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