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15 December, 20:28

If you push a crate horizontally with 100 N across a 10-m factory floor and the friction between the crate and the floor is a steady 70 N, how much kinetic energy does the crate gain?

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Answers (2)
  1. 15 December, 20:29
    0
    300 J.

    Explanation:

    kinetic Energy: This can be defined as the energy of a body by virtue of it's motion. The S. I unit of kinetic energy is Joules (J).

    From the question, Applying the law of conservation of energy,

    Ek = F'*d ... Equation 1

    Ek = kinetic energy of the crate, F' = resultant force on the crate, d = distance moved by the crate.

    But,

    F' = F-Fr ... Equation 2

    Where F = Horizontal force on applied to the crate, Fr = Friction force

    Substitute equation 2 into equation 1

    Ek = (F-Fr) d ... Equation 3

    Given: F = 100 N, Fr = 70 N, d = 10 m.

    Substitute into equation 3

    Ek = (100-70) 10

    Ek = 30 (10)

    Ek = 300 J.
  2. 15 December, 20:32
    0
    300J

    Explanation:

    Parameters given:

    Force applied on crate, F = 100N

    Distance moved by crate, D = 10m

    Frictional force, Fr = 70N

    The work done by pushing the crate is given by:

    W = F * D

    = 100 * 10 = 1000J

    The work done by friction is given as:

    W (Fr) = - Fr * D

    = - 70 * 10 = - 700J

    Work done by friction is negative because it works against the motion of the crate.

    Net work done = 1000 + (-700)

    WD = 1000 - 700 = 300J

    Work done is equivalent to the change in kinetic energy.

    WD = ∆KE

    ∆KE = KE (final) - KE (initial)

    KE (initial) = 0

    Therefore,

    WD = KE (final)

    => KE (final) = 300J
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