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20 February, 17:51

A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2. How big is the frictional force on the block? What is the coefficient of friction between the chair and floor?

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  1. 20 February, 19:26
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    We can first calculate the net force using the given information.

    By Newton's second law, F (net) = ma:

    F (net) = 25 * 4.3 = 107.5

    We can now calculate the frictional force, f, which is working against the applied force, F (app) (this is why the net force is a bit lower):

    f = F (net) - F (app) = 150 - 107.5 = 42.5 N

    Now we can calculate the coefficient of friction, u, using the normal force, F (N):

    f = uF (n) - - > u = f/F (N)

    u = 42.5/[25 (9.8) ]

    u = 0.17
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