A block of mass 1.01 kg is placed on top of a second block of mass 7.38 kg. The

Question

A block of mass 1.01 kg is placed on top of a
second block of mass 7.38 kg. The coefficient
of kinetic friction between the 7.38 kg block
and the surface is 0.26. A horizontal force F
is applied to the 7.38 kg block.
Calculate the magnitude of the force neces-
sary to pull both blocks to the right with an
acceleration of 3.64 m/s2. Assume no slipping
between the two blocks. The acceleration of
gravity is 9.8 m/s2 .
Answer in units of N

Find the minimum coefficient of static friction
µt between the blocks such that the 1.01 kg
block does not slip under an acceleration of
3.64 m/s2.

Explanation / Answer

a. FOrce = (m+M) x a = (7.38 +1.01) x 3.64 = 30.5396 Newton

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