# Homework Help: What's the acceleration of the 2kg block?

1. Jun 16, 2009

### bodensee9

Hello:

A picture is attached. I have 2 blocks that are attached to a horizontal pulley. One block (of 1Kg mass) is lying on top of another (of 2kg mass), and both are attached to two ends of the pulley (see attachment). The lower block is pulled by a rope with a tension force of 20N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient between the lower and upper block is also 0.30. What's the acceleration of the 2kg block?

Wouldn't we have to find out if the first block will slip or not? So then friction max = 9.8*1*0.3 = 2.94, which gives an acceleration max of only 2.94. So if we pulled with 20N - friction on lower block, we have 14.12, we could still make the top block slip. I think the top box will need to slip, because if it didn't, then the friction would need to be greater than tension, meaning that the maximum tension would be equal to the maximum of friction, or 2.94. So 20 - friction on lower block - 2.94 = 11.18, which would accelerate the lower by
5.59 m/s^2. Seems odd to me. I'm not sure how to determine the relationship between thee acceleration of the lower and upper blocks.

The equations are:
T - 1.0*9.8*0.3 = 1.0*a.
20 - 2.0*9.8*0.3 - T = 2.0*(a of the lower box).

Thanks.

#### Attached Files:

• ###### pulley.pdf
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2. Jun 16, 2009

### glueball8

acceleration of top and bottom is the same since the strings length is constant.

For the lower box the bottom friction is mg*Mu note m is not 2 and then there's also the friction on the top.

3. Jun 16, 2009

### bodensee9

so you are assuming that the top block isn't sliding?

Thanks.

4. Jun 16, 2009

### glueball8

The top is sliding, but the first equation seems right.

5. Jun 16, 2009

### bodensee9

so the mass would be box 1 + box 2, or 3 then? I thought that only happens if the top box isn't sliding. thanks.

6. Jun 16, 2009

### glueball8

3 since it is also contributing to the normal force on the ground.