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PHY 201
Your 'cq_1_21.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A typical automobile coasts up a typically paved incline, stops, and coasts back down to the same position.
• When it reaches this position, is it moving faster, slower or at the same speed as when it began? Explain
answer/question/discussion: ->->->->->->->->->->->-> :
‘dPE will be the same but opposite sign because the ‘dy will be the same, just positive and negative. ‘dPE = m*g*’dy and m and g do not change.
‘dW_nc will be the same. Friction is a coefficient * m * g and none of these will change but it will always be negative.
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Actually friction is equal to the coefficient times the normal force, the normal force having magnitude m g only when the surface is level.
Friction isn't the same in both directions, since it reverses direction, but it does the same work in both directions so the train of your reasoning is OK.
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On the way up we have -‘dW_nc_on = ‘dPE - ‘dKE. Since ‘dPE doesn’t change, I solved this for ‘dPE and got ‘dPE = ‘dW_nc - ‘dKE
On the way down, we have -‘dW_nc_on = -‘dPE + ‘dKE. Solving for -‘dPE, -‘dPE = ‘dW_nc + ‘dKE.
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You could make this clearer by using `dPE_up and `dPE_down, which each being the negative of the other.
You could also use `dKE_1 and dKE_2 for the KE changes.
`dW_nc is indeed the same both ways.
`dW_nc_on = `dPE + `dKE in all cases. You can't have -`dW_nc_on = `dPE - `dKE and -`dW_nc_on = -`dPE + `dKE.
As I've said, use of subscripts could reconcile this.
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I am not sure the above makes total sense, I might be wrong. Another way to look at it is summing the forces.
On the way up you would have, assuming down the incline to be positive, F_net = F_g_parallel + F_f, so a = (F_g_par. + f_f)/m, v_0 = ‘sqrt((2*’ds(f_g_para + f_f)/m).
On the way down you would have F_net = F_g_par - F_f so a = (F_g_par. -F_f) and v_f = ‘sqrt((2*’ds(f_g_para - f_f)/m).
Since f_g_para - F_f will obviously be smaller than F_g_par + F_f, the velocity at the bottom on the way down will be less than on the way up.
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Your thinking is good, but you are trying to get too much mileage out of your notation. Check my note.
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