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26 September, 08:25

A person pushes horizontally with a force of 180. N on a 67.0 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.14. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration?

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  1. 26 September, 12:17
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    a) frictional force = 93.8 N

    b) acceleration = 1.29 ms∧-2

    Explanation:

    The frictional force can be calculated by:

    μ k = Fk / N

    where:

    μ k = coefficient of kinetic friction = 0.14 N = normal force or the force perpendicular to the contacting surface Weight of the crate, W = mg = (67 * 10) = = 670 N

    take acceleration due to gravity = 10m/s∧2

    The normal force, N = W = 670N (upward force equals downward force)

    Fk = μ k * N

    = 0.14 * 670

    Fk = 93.8 N

    b) To calculate the magnitude of crate's acceleration

    F - Fk = ma

    where

    Fk = frictional force = 93.8N F = horizontal force = 180 N m = mass = 67.0 kg a = magnitude of acceleration (unknown)

    by substituting,

    180 - 93.8 = 67a

    86.2 = 67a

    acceleration, a = (86.2 : 67)

    a = 1.29 ms∧-2
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