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 = 0.80 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?

______N

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

______m/s² (2.0 kg block)
______m/s² (4.0 kg block)

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

______N (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.

______m/s2 (2 kg block)

______m/s2 (4 kg block)

Answer 1

m2 = 2kg
m4 = 4kg.

Since there is no relative motion between m2 and m4 we use s =0.8.

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 = 0.8*6*9.8 = 47.04 N
--------------------------------------...

b)
a = force / mass = (0.5*47.04) / 6 = 3.92 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 = m_1*a = 2*3.92 m/s^2 = 7.84 N
--------------------------------------...

c)
If the force is 2*47.04 = 94.08 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 = (94.08 -3.92)

Acceleration of 4 kg mass = (94.08 -3.92) / 4 =22.54 m/s^2