• Support PF! Buy your school textbooks, materials and every day products Here!

Varying coefficient of friction

  • Thread starter eeriana
  • Start date
  • #1
15
0

Homework Statement


A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at .100 at P and increases linearly with a distance past P, reaching a value of .600 at 12.5m past point P. a) Use the work energy theorem to find how far this
box slides before stopping.


Homework Equations


Work = .5mV1-.5mV2
K= .5mv^2



The Attempt at a Solution


I am not really sure where to begin, if someone could steer me in the right direction, I would be grateful. I know that the friction area will cause a deceleration until the box stops. I know that the net force of Fx will be the F force in the positive x direction minus the Friction force in the negative x direction. I feel like I need the mass of the box and since it is not given, it must not be necessary and that means I am missing something.
I believe the s vector will be 12.5 since that is the distance of the friction area.

Thanks..

Amy

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
18,821
2,009
Since the friction force is related to the coefficient of friction [itex]\mu[/itex], one can use the fact that the kinetic energy is dissipated by the friction over the distance from P to the point that it stops, where the kinetic energy is of course zero. The friction force is simply [itex]\mu[/itex](x)*mg.

So the initial KE = 1/2 mv2 = energy dissipated (or work done by friction). Note the mg is constant, but the friction coefficient is a function of x or the distance traveled.
 
  • #3
11
1
Alright so I did relate that Work is equal to the change in kinetic energy which is also equal to the magnitude of Force times the magnitude of the change in distance.


By doing so, I get that the initial Kinetic Energy equals mg(0.04x + 0.1)x


The equation for u(x) = 0.04x + 0.1. The distance travelled during the process is x.


When I solve for x, I get the answer 3.984 m, but this seems to be the wrong answer. Does anyone see where I went wrong?
 
  • #4
11
1
bump*
 

Related Threads on Varying coefficient of friction

Replies
5
Views
1K
Replies
20
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
720
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
6
Views
3K
Replies
6
Views
19K
  • Last Post
Replies
5
Views
751
  • Last Post
Replies
2
Views
1K
Top