We just did a lab in Physics titled \"Work and Energy\", and we basically took a

Question

We just did a lab in Physics titled "Work and Energy", and we basically took a weight of different masses (m1 and m2) and attached it to a car(mass of car = 0.461 kg) on an incline of 30 and 45 degrees. The distance the car traveled is 0.302 m, which is also the change in y for the weight suspended in the air. We calculated the work of friction by formulas given in the book (f = mcg sin (30 or 45) - m2g) and Wf= fd. That's all good. What I'm having issues with are the following questions:

1) What was the work done by the suspended weight when the car (a) moved up the incline and (b) moved down the incline?

2) What was the work done by gravity acting on the car when it (a) moved up the incline and (b) moved down the incline?

3) a) for the car going up the incline, what percentage of the work done by the suspended weight was lost to friction? (b) For the car moving down the incline, what percentage of the work done by gravity was lost to friction?

I've already tried working these out, but it just doesn't make sense. Help please!

Explanation / Answer

m1
m2
Mc = 0.461 kg a1 = 30 deg a2 = 45 deg
d = 0.302 m ______________
You said: f = Mc g sin(a) - m g Wf= fd.
That's correct - but only when mass m is moving with constant speed. Here why: Newtons second law: Hanging mass - we use first coordinate system to analyze its motion. X-axis is horizontal (orientation of this axis is not important), and y-axis is vertical with positive direction downwards. There are no motion and forces in x-axis so we don't write any equations for it: Y: m a = m g - T (eq.1) Car - we use different coordinate system for easier analasis. X-axis is parallel with incline with positive direction up the incline, and y-axis is normal on it, directed upwards, away from the incline: X: Mc ax = T - Mc g sin(a) - f (eq.2)    Y: Mc ay = N - Mc g cos(a) = 0 (eq.3) ----> because body is not moving in y-axis at all From eq.1 we get that tension is T = m g - m a. We use this in eq.2 Note that ax is x component of acceleration of car (x component in its coordinate sys.) and it is equal to the acceleration of hanging mass. (Can you prove it mathematically? :D) So using f = Nµ and substituting T: Mc a = (m g - m a) - Mc g sin(a) - f ----> if car is moving downwards sign of f is reversed f = mg - ma - Mc g sin(a) - Mc a f = g (m - Mc sin(a)) - a (m + Mc) (case 1) If the car moved downwards, there would be sign change: f = a (m + Mc) - g (m - Mc sin(a)) f = a (m + Mc) + g (Mc sin(a) - m) (case 2) As you can see. Your formula for f is only valid when a = 0, that is,velocity is constant, and body is moving down the incline (case 2). Work is given by: Wf = f (scalar (or dot) product) d = fd cos (angle between them) This is definition. Actually, teh definition uses integrals, but they can be ommited if force is constant. Therefore there are two cases: 1) Body is moving up the incline and the angle between the displacement and force is zero (note that this angle doesn't depend on our choice of coordinate system but only on force and displacement vectors), therefore: Wf = fd cos0= fd (case 1) 2) Body is moving down the incline and the angle is now 180 degrees, so: Wf = fd cos180 = -fd (case 2) Your formula for f is right if there was no acceleration in experiment. Since you have used it on your class, I assume that there really was no acceleration (a very special case to be precise), so we will use a=0. Now we can answer your questions: 1) What was the work done by the suspended weight when the car (a) moved up the incline and (b) moved down the incline?
Suspended weight did the work W = Fd (F is force with which suspended weight pulls the car, in our case, that would be tension T). a = 0 so T = m g, and W = m g d If car is movinf up the incline sign is positive, and in second case it is negative.   
Work done by gravity is Wg = Fg d cos(angle between them) or Wg = (component of force of gravity parallel with displacement) * d First approach turns to be the same as second after some use of trigonometry. Wg = Mc g sin() d In second case, there is just one minus sign in front of the result. 3) a) for the car going up the incline, what percentage of the work done by the suspended weight was lost to friction? (b) For the car moving down the incline, what percentage of the work done by gravity was lost to friction? You can obtain these results by dividing the Wlost with Wput. In both cases Wlost is work of friction. In case 1 Wput is work of suspended weight, in case 2 Wput is work of gravity. You have the formulas for Wf, W and Wg so it should be no problem to you to find what are you asked for. Peace be upon you. :)

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