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9 December, 23:10

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at a force of 34.8 N. He then maintains this 34.8 N force while the box accelerates at 0.357 m/s^2. What is the coefficient of static friction between the crate and the floor?

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  1. 9 December, 23:19
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    The coefficient of friction between the crate and the floor = 0.143

    Explanation:

    Frictional Force: This is the force that act between two surface in contact and tend to oppose their motion. it is measured in Newton (N)

    F - F₁ = ma ... Equation 1

    Where F = Force of the crate, F₁ = frictional force, m = mass of the crate, a = acceleration of the crate

    making F₁ the subject the equation 1

    F₁ = F - ma ... Equation 2

    Given: F = 34. 8 N, m = 19.8 kg. a = 0.357 m/s².

    Substituting these values into equation 2

    F₁ = 34.8 - (19.8*0.357)

    F₁ = 34.8 - 7.07

    F₁ = 27.73 N.

    F₁ = μR ... equation 3

    making μ the subject of formula in equation 3

    μ = F₁/R ... Equation 4

    Where F₁ = Frictional Force, μ = coefficient of static friction, R = Normal reaction.

    R = mg,

    where g = 9.8 m/s², m = 19.8 kg

    R = 19.8 (9.8) = 194.04 N,

    R = 194.04 N, F₁ = 27.73 N

    Substituting these values into equation 4

    μ = 27.73/194.04

    μ = 0.143.

    therefore the coefficient of friction between the crate and the floor = 0.143
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