A 12 kg block is placed on top of a 30 kg block (see figure below). A force of 3

Question

A 12 kg block is placed on top of a 30 kg block (see figure below). A force of 364 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 0.40. (Assume the positive direction is to the right.)

(a) What is the acceleration of the upper block? (Indicate the direction with the sign of your answer.)

(b) What is the acceleration of the lower block? (Indicate the direction with the sign of your answer.)

(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 block?

Explanation / Answer

a)The only force acting on m2 is the friction force, so from Newton’s 2nd law:

F = m2a1 = µm2g
a1 = µg
= (0.2)(9.81m/s²)
= 1.962m/s² (-x direction)

b) The forces acting on the lower block is the applied force (F) and the friction force between it and the upper block and friction between it and floor:

F = m1a2 = F - µm2g - µfm1g
a2= 7.424m/s2 (+x direction)

c) In that case

m2a2=µsm2g

or µs= a2/g=0.7568

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