Two blocks stacked on one another

[pureouchies4717]

hey guys, i was wondering if you could please help me

question:

The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F_vec causes both blocks to cross a distance of 5.0 m, starting from rest.What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

this is a very tricky question

force of friction for lower block: Ff= 13.72N

Fsmax= 23.52

so the force being exerted has to have a max of 9.8N

F=ma
9.8= 7a
a=1.4 m/s^2

i have no idea how i can get the time from this

 

[Hootenanny]

could you post some of your working / thoughts?

 

[pureouchies4717]

so here, i tried to incorporate it into the kinematics equations

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a1.gif 5=.5(a)t^2
and got a time of 2.67s, which is wrong

 

[Integral]

Where did the friction enter into the problem?

You have jumped a step to far in your choice of equations. Start from F=ma, you should have something in your text relating the friction to this fundamental starting point. You have 2 friction problems to do here. The first will tell you the maximum acceleration you can have without the top block slipping. Then that acceleration and the friction between the floor and the bottom block will give you the time you need.

 

[pureouchies4717]

thanks for the response

so

a=fmax/m
= 5.88m/s^2
d= .5at^2
5= .5(5.88)(t^2)
t= 1.304 (wrong)

 

[Hootenanny]

Consider the maximum force applicable to the blocks : $F = 0.6 \times 4 = 2.4$. Now apply this to the body of two block (m = 7) to give you the maximum force, which will allow you to calc. the acceleration.

 

[pureouchies4717]

guys, thanks a lot for helping

ok so f=ma

2.4=7a
a= .349 m/s^2

d=.5at^2
t= 5.4s (this comes out to the wrong answer)

sorry to bother you guys..

 

[Hootenanny]

You forgot to factor in the force of friction during movement.

 

[pureouchies4717]

ah yes! thanks sooooo much

it came out to 1.75

:!)

 

[kmt271]

Hey, I have the same question but with different values, I'm trying to do it using this method and it just won't work.
My values are:
Box on top m=2.73kg
Box on bottom m=1.41kg
Co-efficient static = 0.641
Co-efficient kinetic = 0.116
Distance = 4.25m

My work:
Force max = fs = 0.641(9.8)(1.41)
= 8.857N
Force of kinetic friction = 0.116(9.8)(1.41+2.73)
= 4.706N
Fnet = 8.857 - 4.706
= 4.15N

4.15 = 4.14a
a = 1.00m/s/s
d = 0.5(1)t^2
t = 2.91 s - Wrong!

Can anyone spot the bugger?

 

[pointguard 2]

"You forgot to factor in the force of friction during movement. "

Can you please explain this?

 

[OleWik]

Yeah, bump! hehe, id like to know the last "formula" there. Thanks

 

[NascentOxygen]

My work:
Force max = fs = 0.641(9.8)(1.41)

Check this.

 

[NascentOxygen]

Yeah, bump! hehe, id like to know the last "formula" there. Thanks

Where do you mean "there"?

 

[jdsp23]

ah yes! thanks sooooo much

it came out to 1.75

:!)

How did you get this ? :S I've tried everything.
Please explain to me how you did this.

 

[jdsp23]

ah yes! thanks sooooo much

it came out to 1.75

:!)

How did you get to this? How did you factor the kinetic friction ?
i don't understand what just happened...

 

[NascentOxygen]

ah yes! thanks sooooo much

it came out to 1.75

How did you get this ? :S I've tried everything.
Please explain to me how you did this.

Hi jdsp23. Welcome to physics forums.

The message to which you refer has a date on it indicating it was posted in 2006. Its author, nick727, has not logged into his physics forums account here for 3 years. It's likely he no longer reads this forum, so your expectations should not be high of receiving a reply from him.

 

[stagename]

And what is it is asking to cross a distance "d" but without providing one?