Calculating kinetic friction?

Question

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.41 m/s. The mass stops a distance S2 = 1.9 m along the level part of the slide. The distance S1 = 1.07 m and the angle q = 33.30°. Calculate the coefficient of kinetic friction for the mass on the surface.

Answer 1

At the top of the inclined plane, the mass has potential energy (U = m*g*h_initial) and kinetic energy (1/2*m*v^2). Once the block stops, it no longer has kinetic energy (since v=0), but it still has potential energy (U = m*g*h_final). The difference in between final potential energy and initial potential energy will be U_final - U_initial, or m*g*(h_final - h_initial). The change will be negative, since it's losing potential energy (as it's losing height).

Now, although I can't see the diagram it alludes to, I think you'll probably have to use a little trig to find the height of its starting position (you can use the height at the level portion as your reference, letting the final height be zero). Since you're finding the difference in height, it really doesn't matter which you use as your reference, so long as you keep that same reference to calculate all applicable heights.

Recall that the change in potential energy plus the change in kinetic energy is equal to the work "done by" the non-conservative forces. (Technically, non-conservative forces can't do work, according to some, but to solve the problem we'll assume they do.) Thus, the sum of your change in potential energy and your change in kinetic energy will equal the work "done by" friction.

Since work is just Force * distance, you can divide out the distance to get just the force. Then, remember that the Force_friction = μ * n, where μ is your coefficient of kinetic friction, and n is your normal force.

Since I can't see the diagram, I'll assume that's everything, but you should use your better judgment to tell if there's anything else in the problem which I might have missed.

Answer 2

Where is the picture?

Assuming that the mass m slides down the incline distance S1 with speed of 2.41m/s and continues to travel distance S2 on the flat/horizontal portion of the path.


We know that force of friction f is the force F that acts on mass m to slow it down.
Force of friction f=u’mg =ma
So a=u’g or
u’=a/g
u’ – is the coefficient we must find

we know that S2=Vt- .5at^2
also V=at (if we move in reverce)
t=V/a
S2=V(V/a) - .5 a(v/a)^2
S2=V^2/a - .5 V^2/a= .5 V^2/a

a=V^2/ (2 S2 )

Finally

u'= V^2/ (2 g S2 )

Just substitute the numbers
u'= (2.41)^2/ (2 x 9.81 x 1.9 )
u’= 0.156

Answer 3

typical reaction ---> 38x9.8 N frictional tension F balances the pulling tension struggling with field from accelerating. coefficient of friction ---> frictional tension/ typical reaction ---> 194/(38x9.8)--->0.fifty two