A crate is pushed 1.50m horizontally with a force of 2.40N under an opposing force of friction of 0.600N. What is the NET work done on the c

Question

A crate is pushed 1.50m horizontally with a force of 2.40N under an opposing force of friction of 0.600N. What is the NET work done on the crate?

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Ladonna 5 years 2021-08-21T03:48:56+00:00 2 Answers 29 views 0

Answers ( )

  1. Philomena
    0
    2021-08-21T03:50:12+00:00
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    Answer:

    2.7 J

    Explanation:

    Given that

    Distance pushed through, s = 1.5 m

    Force of the push, F(push) = 2.4 N

    Opposing force of friction, F(f) = 0.6 N

    See attachment for solution

    The net work done is a sum of the calculated work done, and is given as

    W(net) = 3.6 +(-0.9) + 0 + 0

    W(net) = 2.7 J

    Thus, the net work done is 2.7 J

  2. binhan
    0
    2021-08-21T03:50:22+00:00
    Reply

    Answer:

    2.7 J

    Explanation:

    From the question,

    The Net work done by the crate = Work done by the crate in moving if through a distance- work done against friction.

    W = Fd-F’d……………….. Equation 1

    Where F = force applied to the crate, d = distance moved by the crate, F’ = force of friction

    Given: F = 2.4 N, F’ = 0.6 N, d = 1.5 m

    Substitute these values into equation 1

    W = 2.4(1.5)-0.6(1.5)

    W = 3.6-0.9

    W = 2.7 J

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