You are lowering two boxes, one on top of the other, down the ramp shown in the

Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.460, and the coefficient of static friction between the two boxes is 0.806.

a) What force do you need to exert to accomplish this?

b) What is the magnitude of the friction force on the upper box?

c) What is the direction of the friction force on the upper box?
- up the ramp
- down the ramp

Explanation / Answer

similar problem You are lowering twoboxes, one on top of the other, down the ramp shown in the figure (Intro 1 figure) by pulling on a ropeparallel to the surface of the ramp. Both boxes move together at aconstant speed of Intro 1 figure) by pulling on a rope parallel to the surface of the ramp. Bothboxes move together at a constant speed of 15.0 . The coefficient of kinetic friction between theramp and the lower box is 0.405, and the coefficient of staticfriction between the two boxes is 0.777 We know from the relation v2 - u2 = 2aS Here,u = 0 or a = (v2/2S) Here,v = 15.0cm/s = 15.0 * 10-2m/s and S =[(4.75)2 + (2.50)2]1/2 = 28.1m The friction force on the lower box is F1 = µkm1a Here,µk= 0.405,m1 = 48.0 kg The force on the upper box is F2 = µsm2a Here,µs = 0.777 and m2 = 32.0kg The magnitude of the friction force on the upper box is F = F1 - F2

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