Acceleration of a cart being pulled by a falling mass

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jumbogala
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Homework Statement


I'm doing a lab and getting very confused.

A cart is on a table. It is attached to a string which goes over a pulley. The pulley is on the edge of the table.

On the end of the string there is a mass. When the mass falls straight down, it pulls on the string and makes the cart accelerate too.

Homework Equations


Mass of the cart = 1 kg (constant)
Trial 1: mass falling down: 0.1 kg & mass added to cart: 0.4 kg
Trial 2: mass falling down: 0.2 kg & mass added to cart: 0.3 kg
Trial 3: mass falling down: 0.3 kg & mass added to cart: 0.2 kg
Trial 4: mass falling down: 0.4 kg & mass added to cart: 0.1 kg

The Attempt at a Solution


I am supposed to graph the acceleration of the cart vs. the pulling force. However, I am not sure what this graph should look like.

For trial 1, the pulling force would be (0.1 kg)*(9.81 m/s2)=0.981 N, right? And the theoretical acceleration would be 0.981 N/(0.4 kg+1kg) = 1.962 m/s2.

However, if I graph a vs. F, I don't get a straight line, because the mass of the moving object is always changing. Is this right? In the book they make a big deal about the total mass of the system being constant, so I'm wondering if that has something to do with it.
 
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The big deal is not for nothing: it is not only the cart that accelerates, the pulling mass doesn't stay in the same place either! In fact it accelerates just as fast as the cart. And that requires some force too!
[edit] but I see you correctly take that into consideration.
 
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Thank you! I'm not sure what I was thinking. It makes much more sense to divide by the total mass. I did take it into consideration but then forgot to include in in my calculation.

Basically I just found the net force for each trial based on the weight of the falling mass. Then, I calculated the acceleration of the cart using a spark timer. I put the net force on the y-axis and the acceleration on the x, which gives a slope of m (total mass).

Unfortunately I did the plot by hand and I don't have a scanner handy, but it seems to be working out now. Thanks a ton for your help, much appreciated!