- #1
Brodo17
- 18
- 0
A 5kg block is placed on top of a 15kg block that rests on a frictionless table. The surface between the top and bottom blocks is roughened so that there is no slipping between the blocks. A 40N horizontal force is applied to the top block. What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block?
F=ma
f=mew FN
Ok, in my first attempt I figured that if the blocks don't slide apart, then the entire sysstem would be accelerating. I said 40N = (20)a, and found that a=2m/s^2
Using this I said (5kg)(2m/s^2) = mew FN
10 / (9.8 X 5) = .20
However the answer says that it is actually .6
Then I though that maybe the system could not accelerate at all because the opposing frictional force will stop it from accelerating.
Im not even sure if I am on the right track though, and I cannot figure this out!
Thank you!
F=ma
f=mew FN
Ok, in my first attempt I figured that if the blocks don't slide apart, then the entire sysstem would be accelerating. I said 40N = (20)a, and found that a=2m/s^2
Using this I said (5kg)(2m/s^2) = mew FN
10 / (9.8 X 5) = .20
However the answer says that it is actually .6
Then I though that maybe the system could not accelerate at all because the opposing frictional force will stop it from accelerating.
Im not even sure if I am on the right track though, and I cannot figure this out!
Thank you!