Question

1) A 10 kg lead brick rests on a wooden table. if a force of 46N is required to slide the brick across the table at a constant velocity, what is the coefficient of friction?

2) A horizontal force of 100N is applied to a 200 kg refrigerator sitting on a horizontal surface. the refrigerator remains at rest. what is the value of the frictional force acting on the refrigerator? what will be the value of the frictional force on the refrigerator if the horizontal push is removed?

Show work please so I can see how these type of questions are solved.

Thank you so much!
1. Fb = m*g = 10 *9.8 = 98 N. = ForceFb of brick.

Fap-Fk = m*a
46 - Fk = m*0 = 0
Fk = 46 N. = Force of kinetic friction.

u = Fk/Fb = 46/98 = 0.469.

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