A 2.0 kg block sits on a 4.0 kg block that is on a frictionless table. The coeff

Question

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 = 0.20.

(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 (b) If F is half this value, find the acceleration of each block.

Find the magnitude of the force of friction acting on each block. (2.0 kg block) N (4.0 kg block)

(c) If F is twice the value found in (a), find the acceleration of each block.

Explanation / Answer

m2 = 2kg
m4 = 4kg.

Since there is no relative motion between m2 and m4 we use s =0.30.

There are two frictional forces one due to the normal force m2g = m2g
And another due to the normal force m4g = m4g.
The applied force must be equal and opposite to the sum of these frictinal forces.

a)
F = (m2 + m4) g = 0.20*6*9.8 = 11.76 N

b)
a = force / mass = (0.1*11.76) / 6 = 0.196 m/s^2

The frictionla forces are action-reaction pairs and the frictional force is the same on the two blocks and is given by f = m1 a = 2*0.196 m/s^2 =0.392 N

c)
If the force is 2*11.76 = 23.52 N, then m2 slips on m4 and the frcional force given by

m2a2 = k*m2g = 0.2*2*9.8 = 3.92 and a2 = 0.2*9.8 = 1.96 m/s^2

Net force = (82.32 -3.92)

Acceleration of 4 kg mass = (82.32 -3.92) / 4 =19.6 m/s^2

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