A 10 kg block is placed on top of a 29 kg block (see figure below). A force of 3

Question

A 10 kg block is placed on top of a 29 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 block (accelerating less than the lower block). The coefficient of kinetic friction. between the upper block and the lower block is 0.2, the coefficient of friction between the lower block and the floor is 0.49. What is the acceleration of the upper block? What is the acceleration of the upper block? How big would the coefficient of static friction between the upper and lower block have to be so that upper block not slip on the lower block?

Explanation / Answer

friction force between upper and lower block=friction coefficient*normal force

=0.2*10*9.8=19.6 N

this friction force is acting on upper block from left to right and on lower block from right to left.

so acceleration of upper block=19.6/10=1.96 m/s^2. from left to right

b)friction force between lower block and ground=0.49*(10+29)*9.8=187.278 N

then acceleration of lower block=(364-187.278-19.6)/29=5.418 m/s^2

c)for both the blocks to move simultaneously, the net acceleration of the system=(364-0.49*(10+29)*9.8)/(10+29)=4.531 m/s^2

hence friction force on upper block=mass*acceleration=45.31 N

==> friction coefficient*10*9.8=45.31

==> friction coefficient=0.462

hence friction coefficient has to be 0.462 in order to upper block not slip on lower block.

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