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29 December, 08:44

A 1200-N crate rests on the floor. How much work is required to move it at constant speed (a) 5.0 m along the floor against a friction force of 230 N, and (b) 5.0 m vertically?

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  1. 29 December, 11:04
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    (a) The work required to move the crate = 7150 J

    (b) The Work required to move the crate vertically = 6000 J

    Explanation:

    (a) Frictional Force: These is the force that tend to oppose the motion of two bodies in contact.

    Work required to move the crate = work done against friction + work required to move it through a distance of 5.0 m.

    Wr = Wf + Wd

    Where Wr = work required to move the crate, Wf = work done against friction, Wd = work required to move the crate through 5.0 m

    Given: Wf = friction force * distance = 230*5 = 1150 J

    and Wd = force * distance = 1200*5.0 = 6000 J.

    Substituting these values into equation 1

    Wr = 6000 + 1150

    Wr = 7150 J

    Therefore the work required to move the crate = 7150 J

    (b)

    Work required to move the crate vertically = mgh

    or

    Work required to move the crate vertically = Wh ... Equation 2

    Where W = weight of the block, h = vertical height

    Given:W = 1200 N, h = 5.0 m

    Substituting these values into equation 2

    Work required to move the crate vertically = 1200*5

    Work required to move the crate vertically = 6000 J
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