I have a problem I need help with. I understand part 1, but am including it for

Question

I have a problem I need help with. I understand part 1, but am including it for part 2.

Part 1 - A student wants to slide a steel 15 kg mass across a steel table. What force must the student apply in order to start the box moving? The answer is 108.8N.

Part 2 - What force must the student apply to keep the mass in part 1 moving at a constant velocity (to overcome the kinetic friction between the mass and the table)? The answer is 83.3 but I do not know how to get the answer.

Can someone show me the steps to get the answer for part 2? Thanks!

Explanation / Answer

m =15 kg

for steel coeffecient of static friction us =0.74

for steel coeffecient of static friction uk =0.5668

Part 1: F =Fs = usmg = 0.74*15*9.8

F =108.8 N

Part 2: For moving with constant velocity , applied force should be kinetic fricition.

From Newtons second law

F-Fk =ma =0

F =Fk =uk mg = (0.5668)(15*9.8)

F = 83.3 N

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