A 4.0 kg block is put on top of a 5.0 kg block. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 8 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
(b) What is the resulting acceleration of the blocks?
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
(b) What is the resulting acceleration of the blocks?
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Start with the stationary condition:
Sum of the forces on the top block in horizontal direction = 0 = 8N - friction
friction = 8 N
Now, when the blocks are moving
Sum of the forces on the top block in horizontal direction = m*a = friction
a = friction / m = 8 N / 4.0 kg = 2.0 m/s <--- (b)
This is the acceleration needed to produce 8N 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 + 5.0 kg) * 2.0 m/s^2 = 18.0 N <---- (a)
Sum of the forces on the top block in horizontal direction = 0 = 8N - friction
friction = 8 N
Now, when the blocks are moving
Sum of the forces on the top block in horizontal direction = m*a = friction
a = friction / m = 8 N / 4.0 kg = 2.0 m/s <--- (b)
This is the acceleration needed to produce 8N 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 + 5.0 kg) * 2.0 m/s^2 = 18.0 N <---- (a)