An individual is pushing a 54 kg table. The static coefficient of friction betwe

Question

An individual is pushing a 54 kg table. The static coefficient of friction between the desk and the floor is 0.55, and the kinetic coefficient of friction is 0.51. 1) What is the acceleration of the desk if 492 N of pushing force is applied when the floor is a flat horizontal surface? 2) What is the acceleration of the desk if 492 N of pushing force is applied when the floor has an upward slope of 6°? 3) What is the acceleration of the desk if 492 N of pushing force is applied when the floor has a downward slope of 6°?

Explanation / Answer

part a )

492 N is greater than static friction force

fsk = mu_s*m*g = 0.55 * 54 * 9.8 = 291.06 N

F - mu_k*m*g = ma

492 - 0.51 * 54 * 9.8 = 54 * a

a = 4.113 m/s^2

part b )

in y direction

N - mg costheta = 0

in x direction

F-mgsintheta + fs = ma

fs = mu*N

F-mg(sintheta+mucostheta) = ma

a = 2.726 m/s^2

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