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Old Nov 9th 2009, 03:37 PM   #1
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calculating coefficient !



When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 1.87 m/s. The mass stops a distance S2 = 2.15 m along the level part of the slide. The distance S1 = 1.25 m and the angle theta = 37.1 degrees. Calculate the coefficient of kinetic friction for the mass on the surface.


E = K+U = (0.5)mv^2 + mgS1sin(theta)
m cancelled?
= 9.15

W1 = [mgcos(theta)-uk mgsin(theta)]S1
W2 = -uk mgS2

uk = [ E + mgcos(theta)S1 ]/[mgS2 + mgsin(theta)S1]


m cancelled out

so

uk = [ 9.15 + (9.81)(1.25)cos37.1 ] / [ (9.81)(2.15)+(9.81)(1.25)sin37.1 ]
= 0.664

but wrong...

where did i go wrong?

Last edited by christina; Nov 9th 2009 at 03:39 PM.
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Old Nov 10th 2009, 01:55 AM   #2
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The total initial energy say E gets completely converted into work against friction on the slope and on the level surface.
E = K+U = (0.5)mv^2 + mgS1sin(theta)
m cancelled?
= 9.15

m does not cancel yet. leave it as it is.

The work done against friction on the slope = W1 = force of friction x S1

= uk mg cos theta x S1 = W1.

E = W1 + W2.

Now m will cancel out. Solve for uk.
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Old Nov 10th 2009, 08:38 AM   #3
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so am i right about the W2 ?
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Old Nov 10th 2009, 11:49 PM   #4
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Yes . However, in

W2 = -uk mgS2

you have to remember that the work is a dot product between force and displacement giving F s cos 180 = - F s in our case. Thus the minus sign
cancels out in W2 and W1 also is to be reated as +ve, i.e., when you add the two,

W1 + W2 = uk mg cos theta x S1 + uk m g S2
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