Ask Question
14 January, 06:15

A block weighing 8.5 N requires a force of 2 Nto push it along at constant velocity. What is the coefficient of friction for thesurface?

+2
Answers (1)
  1. 14 January, 06:32
    0
    The coefficient of friction of the surface μ = 0.235

    Explanation:

    Values of the given variables:

    Weight of block = 8.5 N

    Pushing force F = 2 N

    velocity = constant

    Solution:

    Newton's Second Law of motion states that acceleration of an object depends on the mass of the object and the force applied

    F = m a

    Since v = constant, a = 0. Thus

    Net force Fnet = 0.

    Now F_net is the sum of the pushing force and the frictional force along the horizontal, which opposes the motion:

    Fnet = Fpushing - Ffrictional_force

    Given that Fnet = 0, Fpushing = Ffrictional_force. and Ffrictional_force = 2 N

    Equation of friction is given by

    Ffrictional_force = μ*W

    2 = μ*8.5

    μ = 2/8.5 = 0.235
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A block weighing 8.5 N requires a force of 2 Nto push it along at constant velocity. What is the coefficient of friction for thesurface? ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers