Physics Help Forum calculating coefficient !

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 Nov 9th 2009, 02:37 PM #1 Senior Member   Join Date: Sep 2009 Location: St. John's, NL Posts: 124 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 02:39 PM.
 Nov 10th 2009, 12:55 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Nov 10th 2009, 07:38 AM #3 Senior Member   Join Date: Sep 2009 Location: St. John's, NL Posts: 124 so am i right about the W2 ?
 Nov 10th 2009, 10:49 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 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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