A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and between the lower box and the table is 0.600, while the coefficient of static friction between these same surfaces is 0.800. A horizontal pull to the right is exerted on the lower box, as shown in the figure, and the boxes move together. What is the friction force on the UPPER box?

http://session.masteringphysics.com/problemAsset/1427072/2/p5.55.jpg

The answer is 19.3 N to the right but an explanation would be nice.

Question

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and between the lower box and the table is 0.600, while the coefficient of static friction between these same surfaces is 0.800. A horizontal pull to the right is exerted on the lower box, as shown in the figure, and the boxes move together. What is the friction force on the UPPER box?

http://session.masteringphysics.com/problemAsset/1427072/2/p5.55.jpg

The answer is 19.3 N to the right but an explanation would be nice.

anonymous · Answer

Strategy:
1) Find the total acceleration a=Sigma F/(m1+m2)
2) Find the total force acting on upper block F=m up*a.

anonymous · Answer

hmm still can't seem to find the answer. would sigma F be 150 - the static friction force for the two blocks?

anonymous · Answer

thanks just got it.