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9 January, 04:42

A crate pushed along the floor with velocity v⃗ i slides a distance d after the pushing force is removed. if the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping?

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  1. 9 January, 07:59
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    Define

    m = the mass of the crate.

    F = μmg, the resistive dynamic frictional force,

    where μ = dynamic coefficient of friction.

    a = the deceleration, when the crate slides subject only to to the frictional force.

    The crate travels a distance d with initial velocity of v. Therefore

    v² - 2ad = 0

    a = v² / (2d) (1)

    Also,

    F = ma

    F = (mv²) / (2d) (2)

    When m is doubled, then

    F = (2mv²) / (2d) = (mv²) / d

    The corresponding deceleration is

    a = F/m = v²/d

    Therefore, the new distance traveled, D, is given by

    v² - 2 (v²/d) D = 0

    D = d/2

    The new distance traveled is one half of d.

    Answer: d/2.
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