Okay, so I have a problem on my hw worded like this:
Once a 30 kg crate is in motion on a horizontal floor, a horizontal force of 50 N keeps the crate moving with a constant velocity. The acceleration of gravity is 9.81 m/s^2;
What is the coefficient of kinetic friction between the crate and the floor?
My solution:
I basically took the equation that states that the coefficient of kinetic friction equals the force of the friction divided by the mass times acceleration.
So, I did this: Coefficient= 50/ (30*9.81).
Does this make any sense?
2007-03-15 16:02:21 · 5 answers · asked by Nikita R 2 in Science & Mathematics ➔ Physics
Yes! you are correct
Better yet a little mathematics
f=uN
f -force of friction
u - coefficient of friction
N - normal or perpendicular to the plane force
N=w= mg
m - mass
g - acceleration due to gravity
w - weight of the object.
Have fun!
2007-03-15 16:07:05 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
Absolutely it makes sense!
F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force.
If the crate is moving at a constant velocity, that means friction is equal to the force moving the box, so F = 50N.
The normal force is, in this case, simply the weight of the box, so mass * gravity, or 30*9.81.
Substituting, you get:
50 = (30*9.81)μ
μ = 50/(30*9.81)
Good job! =D
2007-03-15 16:08:09 · answer #2 · answered by cluekoo 4 · 0⤊ 0⤋
Yes, you are simply calculating downward force, that is Mass x acceleration or 30 x 9.8= F
Then the frictional force is always normal downward force F divided into the horizontal force whichis 30 K
50/Fn
2007-03-15 17:56:10 · answer #3 · answered by James M 6 · 0⤊ 0⤋
This makes sense because you'll get around .17 which is a reasonable coefficient of kinetic friction. It sounds like you answered your own question - good job though.
2007-03-15 16:10:25 · answer #4 · answered by Elwood Blues 1 · 0⤊ 0⤋
no
2007-03-15 16:08:21 · answer #5 · answered by Conqi 5 · 0⤊ 1⤋