Ask Question
28 March, 13:07

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?

+2
Answers (1)
  1. 28 March, 14:05
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers