A block of mass m t = 4.0 kg is put on top of a block of mass m b = 5.5 kg. To c

Question

A block of mass mt = 4.0 kg is put on top of a block of mass mb = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 21 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.

A block of mass mt = 4.0 kg is put on top of a block of mass mb = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 23 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table. F that can be applied to the lower block so that the blocks will move together. N (b) Find the magnitude of the resulting acceleration of the blocks. m/s2 (a) Find the magnitude of the maximum horizontal force

Explanation / Answer

Start with the stationary condition:
Sum of the forces on the top block in horizontal direction = 0 = 21N - friction
friction = 21 N

Now, when the blocks are moving
Sum of the forces on the top block in horizontal direction = m*a = friction
a = friction / m = 21 N / 4.0 kg = 5.25 m/s <--- (b)

This is the acceleration needed to produce 21N of force to break the static friction.

Sum of the forces globally on the two blocks together = (m1+m2) * a = F
F = (4.0 kg + 5.5 kg) * 5.25 m/s^2 = 49.875 N <---- (a)

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