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Slope of the graph acceleration versus mass

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I need help!

Soo, I'm doing a lab report about Newton's Second Law and I've encountered a small problem.

I've done the graph of acceleration versus MASS OF WEIGHT (not mass of the trolley but mass of the weight) and I don't know it's slope...

I have trolley which is put on the table which has, at it's end a rotating wheel on top of which goes a string connected to the trolley at one, and a weight at the other side. And I'm calculating how is the acceleration dependant on the mass of the weight pulling it down. (The apparatus is in the picture below.)

And, well, if I go by the Second Newton's Law and write the equation of the system, it would be: mg=(M+m)*a, because the tension cancels itself.

where m is mass of the weight, and M mass of the trolley.

And when I put the acc. on the one side I get : a=mg/(M+m)

So iI though that the gradient of the graph would be --> a=m* g/(M+m), so the gradient is g/(M+m), or???

And what is it at all?

What do I get? :D

Help, please?

post-94536-0-41640000-1333818745_thumb.p

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Can't you put the material you remove from the weight on to the trolley, so that the total mass (M + m) remains constant while you vary the accelerating force mg? Then (assuming zero friction and air resistance, ideal string and massless wheels etc) you could plot a against mg to obtain a straight line of gradient 1/(M+m).

If not, I would suggest plotting a against m/(M+m) in order to obtain a straight line plot with gradient g.

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Thanks :) but I've already done that. The fact is that I need to establish a relationship between acceleration and the mass of the weights pulling the trolley and make a graph out of it. And I got g/(M+m) as a slope... And I got confused, XD. The measuring units are weird m/s^2/kg , so that can't be Newton, because Newton is kg*m/s^2 :D. So I was just wondering whether this slope I got means anything or not... If not, I'll just put that graph down and say that the units don't mean a thing, the slope either, and explain everything on the a vs. m/(M+m) graph :D

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