The only force acting on a 2.0 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 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 8.0 m/s in the positive x direction and some time later has a velocity of 10.0 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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Thông Đạt 4 years 2021-08-25T09:56:15+00:00 1 Answers 212 views 0

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  1. Jezebel
    0
    2021-08-25T09:57:32+00:00
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    Answer:

    W_{net}=36J

    Explanation:

    Given data

    Mass m=2.0 kg

    Magnitude of force F=5.0 N

    Initial velocity Vo=8.0i m/s

    Final velocity Vf=10.0j m/s

    The change in the kinetic energy of canister equals to net work done on the canister

    So

    ΔK=Wnet

    Kf-Ki=Wnet

    For initial kinetic energy

    K_{i}=\frac{1}{2}mv_{i}^2\\K_{i}=\frac{1}{2}(2.0kg)(8.0m/s)^2\\K_{i}=64J

    For final Kinetic energy

    K_{f}=\frac{1}{2}mv_{f}^2\\K_{f}=\frac{1}{2}(2.0kg)(10m/s)^2\\K_{f}=100J

    Work done on canister by 5.0N force is given as:

    W_{net}=K_{f}-K_{i}\\W_{net}=100J-64J\\W_{net}=36J

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