Understanding the Acceleration of a Bureau on a Rough Surface

[musicfairy]

Homework Statement

A bureau rests on a rough horizontal surface (\mu_s = 0.50, \mu_k = 0.40). A constant horizontal force, just sufficient to start the bureau in motion, is then applied. The acceleration of the bureau is:

A. 0
B. 0.98 m/s²
C. 3.3 m/s²
D. 4.5 m/s²
E. 8.9 m/s²

The Attempt at a Solution

The answer is B and I can't figure out why. I tried to use only \mu_s and set friction equal to ma. It gave me C.

I can see where B comes from, kinetic friction is subtracted from static friction. Can someone explain why?

 

[LowlyPion]

Homework Statement

A bureau rests on a rough horizontal surface (\mu_s = 0.50, \mu_k = 0.40). A constant horizontal force, just sufficient to start the bureau in motion, is then applied. The acceleration of the bureau is:

A. 0
B. 0.98 m/s²
C. 3.3 m/s²
D. 4.5 m/s²
E. 8.9 m/s²

The Attempt at a Solution

The answer is B and I can't figure out why. I tried to use only \mu_s and set friction equal to ma. It gave me C.

I can see where B comes from, kinetic friction is subtracted from static friction. Can someone explain why?

The static friction is greater than the kinetic friction. Moving a bureau across a carpet for instance once you get it going you want to keep it going because, once it stops, its harder to get going again.

That's what the problem is telling you. Once the bureau starts - it breaks the Static friction, then the only retarding force is the Kinetic friction. But it is .1 lower. So the surplus .1 goes into accelerating things.

 

[musicfairy]

Thanks a lot. I understand now.