Frictional Force: Find Accel. of 3kg & 2kg Blocks

[ubiquinone]

Question: A force of 29.0 N is applied to a 3.00kg block which is initially at rest on a rough horizontal surface. A 2.00kg block is initially at rest on the 3.00kg block. The coefficient of friction between the blocks is \mu_s=0.35 and the kinetic friction between the 3.00kg block and the table is \mu_k=0.47.
a) Find the acceleration of the 3kg block.
b) Find the acceleration of the 2kg block.

 

[Zeitgeist]

I think, One thing you could do is that, find the force of friction and then find then thr net force, after that since you have the net force and the mass, you can find acceleration.

 

[ubiquinone]

Would it be something like this:
Let the applied force of 29N be F.
a) For the 3.0kg block, the frictional force between it and the table = \mu_k(3.0kg+2.0kg)g=0.470(5g)N
The frictional force between the blocks also act upon the 3.0kg block = 0.38(2g)N
Writing, F_{net} for the 3.0kg block, we have 29N-0.38(2g)-0.470(5g)=3a
Solving for a, we have a=-0.49m/s^2
b) For the 2.0 kg block, the only force affecting its motion is the frictional force between it and the 3.0 kg block. That frictional force = 0.38(2g)N. So F_{net}=0.38(2g)N=2a, solving for a, we have a=3.7m/s^2

Please, correct me if I am wrong, it would help me greatly. Thanks!

 

[OlderDan]

Would it be something like this:
Let the applied force of 29N be F.
a) For the 3.0kg block, the frictional force between it and the table = \mu_k(3.0kg+2.0kg)g=0.470(5g)N
The frictional force between the blocks also act upon the 3.0kg block = 0.38(2g)N
Writing, F_{net} for the 3.0kg block, we have 29N-0.38(2g)-0.470(5g)=3a
Solving for a, we have a=-0.49m/s^2
b) For the 2.0 kg block, the only force affecting its motion is the frictional force between it and the 3.0 kg block. That frictional force = 0.38(2g)N. So F_{net}=0.38(2g)N=2a, solving for a, we have a=3.7m/s^2

Please, correct me if I am wrong, it would help me greatly. Thanks!

There are some errors here. Does it make sense to you that the 3kg block has negative acceleration? Does it make sense that the 2kg has a positive acceleration while the 3kg block has a negative aceleration?

I'm not sure if the .35 you state in the problem or the 0.38 you use in the computation is the correct number. Be sure to check this. Either way, you cannot assume that the actual force of friction between the blocks is the maximum possible force. If the blocks do not slip, the force can be much less than the force you calculated.

 

[ubiquinone]

Alright thanks a lot OlderDan for replying.

I rethought of the problem and may someone please tell me if my logic now is correct.

The static friction between the blocks is 0.35, the kinetic friction between the 3.00kg block and the surface is 0.47, we can find kinetic friction first.

F_{kf}=\mu_k(m_1+m_2)g=0.47(5.00kg)(9.8m/s^2)=23.03N
So the horizontal applied force is 29N-23.03N=5.97N

Now we calculate the static friction between the blocks, if it is greater than 5.97N, then the 2.00kg block on top does not move.

F_{sf}=\mu_s(m_2)g=0.35(2.00kg)(9.8m/s^2)=6.86N. Because 6.86N>5.97N, the 2.00kg block does not move.

So we can treat the two blocks as one massive block with mass of 5.00kg. Then 5.97N=(5.00kg)a and a=1.194m/s^2

Part b) wouldn't the 2.00kg and the 3.00kg mass then have the same acceleration of 1.194m/s^2?

 

[OlderDan]

Alright thanks a lot OlderDan for replying.

I rethought of the problem and may someone please tell me if my logic now is correct.

The static friction between the blocks is 0.35, the kinetic friction between the 3.00kg block and the surface is 0.47, we can find kinetic friction first.

F_{kf}=\mu_k(m_1+m_2)g=0.47(5.00kg)(9.8m/s^2)=23.03N
So the horizontal applied force is 29N-23.03N=5.97N

Now we calculate the static friction between the blocks, if it is greater than 5.97N, then the 2.00kg block on top does not move.

F_{sf}=\mu_s(m_2)g=0.35(2.00kg)(9.8m/s^2)=6.86N. Because 6.86N>5.97N, the 2.00kg block does not move.

So we can treat the two blocks as one massive block with mass of 5.00kg. Then 5.97N=(5.00kg)a and a=1.194m/s^2

Part b) wouldn't the 2.00kg and the 3.00kg mass then have the same acceleration of 1.194m/s^2?

This is the correct conclusion, but the comparison of forces that led you to it is not quite right. The 6.86N is the maximum possible force on the 2kg block, but the 5.97N force is the net force on the combined 3kg+2kg system assuming the 2kg block does not slide. What you should be comparing to verify that the 2kg block does not slide is the maximum possible force to the force required to give the 2kg block the acceleration you computed. The net force on the system could have been greater than 6.86N and the 2kg block still not slide.

I know what you meant, but it is not correct to say the 2kg block does not move. What you mean is that it does not move relative to the 3kg block or does not slide. To an observer watching the blocks react to the applied force, the 2kg block will certainly move.