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

Question

1. 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 = 0.60 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)

2. 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?
? 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)

Please help me answer these 2 questions! They are both the same with different numbers

Explanation / Answer

m2 = 2kg
m4 = 4kg.

Since there is no relative motion between m2 and m4 we use s =0.6.

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 = 0.6*6*9.8 = 41.16 N
--------------------------------------...

b)
a = force / mass = (0.5*41.16) / 6 = 3.47 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*3.47 m/s^2 = 6.94 N
--------------------------------------...

c)
If the force is 2*41.16 = 82.32 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

2)

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

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
--------------------------------------...

c)
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 -1.47)

Acceleration of 4 kg mass = (35.28 -1.47) / 4 =8.45m/s^2

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