Finding coefficient of kinetic friction given mass & horizontal force?

1. Dec 20, 2007

pebbles

1. The problem statement, all variables and given/known data

You use 30N (**Not 25 N, that was a typo) to slide a wooden box that is 12 kg across a floor at constant velocity. What is the coefficient of kinetic fricition?

2. Relevant equations

Ff=coeff of kinetic friction (mu k) m*g
?? not sure.

3. The attempt at a solution

coeff of kinetic friction=12 kg/30N=.4.

That answer is obviously incorrect, but I don't know what to do.

Last edited: Dec 21, 2007
2. Dec 20, 2007

rl.bhat

A body moves across a floor at constant velocity, when you overcome the frictional force. Your formula for frictional force is correct. From where did you get 30 N?

3. Dec 20, 2007

pebbles

oh oops. that was a typo. 25 N should be 30 N.

.4 is correct?

4. Dec 20, 2007

rl.bhat

No. What is the frictional force?

5. Dec 20, 2007

pebbles

hm. i'm looking through my text book and i can't find a formula for frictional force and i can't remember how to get it....a hint please?

6. Dec 20, 2007

pebbles

well, wait Ffriction=muK*m*g.

would it just be 12 kg * 9.8 m/s^2?

7. Dec 21, 2007

Dick

Ffriction=25N. Since that is what's producing the force you are pushing. mg is the normal force. The coefficient of friction, if memory serves, is related to a ratio between those two numbers.

8. Dec 21, 2007

pebbles

Dick, actually, it's not 25N, that was a typo. Question is actually 30 N.

so, Ff=30N-->
coefficient of kinetic friction=30 N/12.0 kg * 9.80 m/s^2= .26.

Am I right?

I thank everyone for their input. :]

9. Dec 21, 2007

Dick

Well, yeah. Ffriction/Fnormal=mu_kinetic. Right?

10. May 10, 2010

hannanm

Maybe use F=MA??

11. Feb 9, 2012

Astro Boy

Hi guys!

for a body at rest the friction acting on it is called static friction! and when the body starts to move it possess Kinetic Friction . The max value to which static friction can increase before the body starts relative motion is called limiting friction.. Hence when a body is placed in the horizontal surface it exerts a force of its own wt downwards which is balanced by the normal reaction force !! N=mg. for dis condition fms=u*N, where u is the coefficient of friction , actually its mew i dnt knw where i can put dt symbol in ma keyboard. bt N=mg so coeff of friction u= fms/mg;

Dis is d main formulae .. i believe dt it would have cleared ur doubts!!

12. Apr 23, 2012

Hi Guys, tried looking for info about the below topic but couldn't find anything. We did an experiment at school ( Fliuds using the TE 64 aparatus) need to conduct a research the friction factor and explain the differences with the pipe coefficient

13. Oct 15, 2013

phisik

Possibility solution

For your prob, I attempted to find the solution and found it is equal to -0.2551020408 ~ -0.26 (if you want to round it into 2-digits)
So this is how I found this answer
Given: 30 N = F ( some forces)
12 kg = m (mass)
so gravity force, F(g) = 12 kg x 9.8 m/s^2 = 117.6 N
Since from the problem said you applied 30 N to slide the wooden box, which there is an arrow showing the box moving forward, but at the same time there is also another force moving backward, equally, as well, friction (Newton 3rd Law).
So F(f) = -30 N
Since F(f) = μ * N => -30 = μ * 117.6 => μ = -30/ 117.6 = -0.2551020408 or -0.26

14. Feb 24, 2015

physicsisnotfun

how would the answer change if the force of friction accelerates at a rate of 3m/s^2

15. Feb 24, 2015

Dick

Forces of friction don't accelerate. You'll have to state the problem much more clearly than that. Also suggest you start a new thread rather than resurrecting an old one.