Friction on lab table In order to make room for an experiment, a physics student

Question

Friction on lab table

In order to make room for an experiment, a physics student is pushing a lab table across the floor. The table has a width of 1.2m, a length of 2.6m, and a mass of 37 kg. The coeffient of static friction between the table and the floor is ?s = 0.18 and the coefficient of kinetic friction is ?k = 0.16.

a) She initially pushes with a force of 55 N. What is the magnitude of the force of friction between the table and the floor?

b) She now pushes with a force of 60 N. What is the magnitude of the force of friction between the table and the floor?

c) She now pushes with a force of 71 N. What is the magnitude of the force of friction between the table and the floor?

d) She now pushes with a force such that the table moves with a velocity of 1.5 m/sec. What is the magnitude of the force of friction between the table and the floor?

e) She now pushes with a force such that the table accelerates at 1.2 m/sec2. What is the magnitude of the force of friction between the table and the floor?

Explanation / Answer

a)

fs,max = s m g = 0.18 * 37 * 9.8 = 65.268 N

55N is less than 65.268N , therefore the table will not move and the frictional force is 55 N

b)

fs,max = s m g = 0.18 * 37 * 9.8 = 65.268 N

60N is less than 65.268N , therefore the table will not move and the frictional force is 60 N

c)

71 N is greater than 65.268 N , therefore the table will move.

fk = k m g = 0.16 * 37 * 9.8 = 58.016 N

d)

the table is moving, therefore

fk = k m g = 0.16 * 37 * 9.8 = 58.016 N

e)

the table is moving, therefore

fk = k m g = 0.16 * 37 * 9.8 = 58.016 N

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