This will look long, but I already have the answers for all but one part. I\'m j

Question

This will look long, but I already have the answers for all but one part. I'm just including all the parts so you have any dependent information for the part I can't answer, which is Part C for the 2.0kg block.

A 2.0 kg block sits on a 4.0 kg block that is on a frictionless table. The coefficients of friction between the blocks are µs = 0.30 and µk = 0.10.

(a) What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slide?
17.64 N

(b) If F is half this value, find the acceleration of each block.
1.47 m/s2 (2.0 kg block)
1.47 m/s2 (4.0 kg block)
Find the magnitude of the force of friction acting on each block.
2.94 N (2.0 kg block)
2.94 N (4.0 kg block)

(c) If F is twice the value found in (a), find the acceleration of each block.
m/s2 (2 kg block) (I tried 17.64-1.96 / 2, but the answer was wrong)
8.33 m/s2 (4 kg block)

Explanation / Answer

c)

here

the force is double therefore F = 2 * 17.64 = 35.28 N

uk * m1 * g = 0.1 * 2 * 9.8 = 1.96

then

m1 * a1 = k * m1 * g

a1 = 0.1 * 9.8 = 0.98 m/s^2

the acceleration for the 2 kg block is 0.98 m/s^2

the net force = (35.28 - 1.96 ) = 33.32 N

acceleration for the 4 kg block is 33.32 / 4 = 8.33 m/s^2

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