A 10 kg block is placed on top of a 28 kg block see figure below). A force of 35

Question

A 10 kg block is placed on top of a 28 kg block see figure below). A force of 358 N is applied to the right on the lower block, and the upper block slips on the lower block (accelerating less than the lower block). The coefficient of kinetic friction between the upper block and the lower block is 0.2, and the coefficient of kinetic friction between the lower block and the floor is 0.50. (Assume the positive direction is to the right.) m2 In (a) What is the acceleration of the upper block? (Indicate the direction with the sign of your answer.) m/s2 (b) What is the acceleration of the lower block? (Indicate the direction with the sign of your answer.) [18.74 ]× m/ (c) How big would the coefficient of static friction between the upper and lower block have to be so that the upper block would not slip on the lower s2 block?

Explanation / Answer

Let’s determine each friction force.

For the 10 kg block, Ff = 0.2 * 10 * 9.8 = 19.6 N

For the floor, Ff = 0.50 * (10 + 28) * 9.8 = 186.2 N

(b)

Net force = 358 – 186.2 = 171.8 N

171.8 = (28+10) * a

a = 4.52 m/s2

(c)

For the upper block to not slip,

force of friction must cancel out the force on the block

That is,

f = F

=> um1g = F

=> u = F/(m1g)

= (358)/(10*9.8)

= 3.65

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