The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of

Question

The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 3.6 m/s in the positive x direction, and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time

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Nho 4 years 2021-09-02T02:45:15+00:00 2 Answers 7 views 0

Answers ( )

  1. yenoanh
    0
    2021-09-02T02:46:27+00:00
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    Explanation:

    Below is an attachment containing the solution.

  2. ngochoa
    0
    2021-09-02T02:46:36+00:00
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    Answer:

    The work done on the canister by the 5.0 N force during this time is

    54.06 Joules.

    Explanation:

    Let the initial kinetic energy of the canister be

    KE₁ = \frac{1}{2} mv_1^{2} = \frac{1}{2} *3*3.6^{2} = 19.44 J in the x direction

    Let the the final kinetic energy of the canister be

    KE₂ = \frac{1}{2} mv_2^{2} = \frac{1}{2} *3*7.0^{2} = 73.5 J in the y direction

    Therefore from the Newton’s first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

    = 73.5 J – 19.44 J = 54.06 J

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