A 12 kg block is placed on top of a 34 kg block (see figure below). A force of 3

Question

A 12 kg block is placed on top of a 34 kg block (see figure below). A force of 352 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 0.47. (Assume the positive direction is to the right.) What is the acceleration of the upper block? (Indicate the direction with the sign of your answer.) What is the acceleration of the lower block? (Indicate the direction with the sign of your answer.) 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 block?

Explanation / Answer

Here,

mass, m1 = 34 Kg

m2 = 12 kg

force ,F = 352 N

coefficient of friction between blocks , u1 = 0.2

coefficient of kinetic friction , u2 = 0.47

a) let the accleration of upper block is a

Using second law of motion

m2 * a = u1 * m2 * g

12 * a = 0.2 * 12 * 9.8

a = 1.96 m/s^2

the acceleration of the upper block is 1.96 m/s^2

b)let the acceleration of the lower block is a

Using second law of motion

34 * a = F - u1 * m2 * g - u2 * (m1 + m2) * g

34 * a = 352 - 0.2 * 9.8 * 12 - 0.47 *(12 + 34) * 9.8

solving for a

a = 3.43 m/s^2

the acceleration of lower block is 3.43 m/s^2

c)

for blocks to not slipping

(34 + 12) * a = 352 - 0.47 * (12 + 34) * 9.8

a = 3.05 m/s^2

let the coefficient of friction is u1

u1 * g = a

u1 * 9.8 = 3.05

u1 = 0.311

the coefficient of static friction between the lower and upper block is 0.311

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