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?

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

__________ m/s^2(2.0 kg block)

__________ m/s^2 (4.0kg block)

Find the magnitude of the force of friction acting on each block.

__________ N (2.0 kg block)

__________ N (4.0 kg block)

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

__________ m/s^2 (2.0 kg block)

__________ m/s^2 (4.0 kg block)

Explanation / Answer

m2 = 2kg , m4 = 4kg.

Since there is no relative motion between m2 and m4 we use s =0.3
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 = 0.3*6*9.8 = 17.64 N

b) a = force / mass = (0.5*17.64) / 6 = 1.47 m/s^2

c) 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*1.47 m/s^2 = 2.94 N

d) If the force is 2*17.64 = 35.28 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 = (35.28 -3.92) =31.36 N
Acceleration of 4 kg mass = (35.28 -3.92) / 4 =7.84 m/s^2

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