# Two boxes on one another

1. Oct 4, 2009

### chris097

The problem statement, all variables and given/known data

The coefficient of static friction is 0.636 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.138. Force F causes both blocks to cross a distance of 7.00 m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.14 kg and the mass of the upper block is 2.80 kg?

http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0832.png [Broken]

The attempt at a solution

Fmax= UsN
=(.636)(2.80*9.81)
=17.4696 N

F(bottom) = UkN
= (.138)((2.80 + 1.14) * 9.81)
=5.3339

Fnet = 17.4696 - 5.3339
=12.1357

F=ma
a=F/m
=12.1357/(1.14+2.8)
=3.0801

d=v1*t + .5at^2
t=square root of (d/(.5a))
= square root of (7/(.5(3.0801)))
=2.13 s

where am i going wrong?
thank you

Last edited by a moderator: May 4, 2017
2. Oct 4, 2009

### willem2

this is the maximum net force that can accelerate the bottom block.
you should consider only the acceleration of the bottom block

3. Oct 4, 2009

### chris097

Thank you.

But what do you mean consider on the bottom block. Do you mean use only the weight of the bottom block?

4. Oct 4, 2009

### willem2

yes. Since you found the maximum net force on the bottom block, then you need to use the mass and the acceleration of the bottom block as well in F = ma