# Homework Help: Acceleration vs Force graphWhat am I doing wrong?

1. Jul 29, 2014

### Crovati

I have to confirm newtons 2nd law via a track and trolley experiment

1. The problem statement, all variables and given/known data

The trolley’s mass is kept constant.
The weight of the cart (252g) plus two bar weights (500g) is = 752 grams, plus additional weights of either, 10g, 15g, 20g, 25g or 30g for the different trials.

The acceleration weights are:
25g
20g
15g
10g
5g
for the different trials.

We used a motion sensor and a software program to get ΔV/Δt.

This is what i got in the end...

Acceleration mass (g)--------Force (N)-----------Average acceleration ms-2

---------5-----------------------0.049---------------------0.10
---------10----------------------0.098---------------------0.22
---------15----------------------0.147---------------------0.36
---------20----------------------0.196---------------------0.44
---------25----------------------0.245---------------------0.63

2. Relevant equations

F=ma
a=(1/m)F

3. The attempt at a solution

I plotted an acceleration vs Force graph, which gave me slope=2.6. And I thought that since F=ma, then the slope would be 1/m, and m= 1/2.6 =0.3846... kg.
If i’m thinking correctly then this should be the value for the mass of the trolley (the cart+the bar weights+the additional weights+acceleration mass)

But when i add all of the components of the trolley’s mass I get 787g...which is nowhere near 0.3846... kg.

I’m not sure what i’m doing wrong, or if I’m going about solving this in completely the wrong way. Or maybe I made an error during the experiment?
I’m completely stuck now and don’t know how I can finish my report. I’de really love some help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 29, 2014

### Bandersnatch

Hi Crovati,

First, let me recap the setup of the experiment, to check if I uderstood it correctly.

There's a mass m allowed to fall under gravity, which is connected via a pulley to another mass M, namely a trolley standing on a flat surface. You measure the acceleration of the system and aim to show that the net force and acceleration are connected by the proportionality factor m+M.

If that's correct, then you've omitted one important force acting on the system. Have you tried drawing a free body diagram? Remember that in $F=ma$ the F is the sum of all forces.
A hint would be: would you measure different acceleration if the trolley were placed on sand, or ice? Or: could mass m be so small that the trolley wouldn't move at all?

3. Jul 29, 2014

### Crovati

I’m not really sure If we’re thinking the same

My instruction for this experiment was to “use the raw data to confirm newton’s second law” which I assumed
(according to F=ma) would be to see if the acceleration of the cart is proportional to gravitational force. The total mass of the trolley (M) plus the acceleration mass (m) are kept constant by transferring masses from the trolley to the hanger for different trials. And the trolley is more of a glider on an air track, so friction is negligible. I should have mentioned that, I guess that’s the extra force you were talking about

The graph I got is linear, so It confirms Newton’s 2nd law, but the value for m+M that I get for the graph is completely different from the one I calculated.

I guess I’m still pretty lost

4. Jul 29, 2014

### olivermsun

Double check the setup of the experiment? You're sure they're 500 g weights and not 50 g? :)

Any other data available for trouble shooting?

5. Jul 29, 2014

### olivermsun

It would be hard to imagine friction causing the acceleration to be too large by about a factor of two.

6. Jul 29, 2014

### Bandersnatch

Damn. You're right, of course.

Whatever the reason for the major discrepancy turns out to be, do note that the slope is not constant. It's 2.6 for the 25g weight only, flattening to about 2 for the 5g weight. Normally that'd suggest friction, or a very low repetition leading to a large error. In the latter case, the extreme value shouldn't be used.

Anyway, since the factor of error is so close to 2, I'll make a wild guess and suggest it was an error during calculation of average values. How did you arrive at the avg. acceleration values - step by step? What was the raw data you actually measured?

7. Jul 29, 2014

### Crovati

I considered that myself, but I checked with my lab partner and we had the same values for weights and everything. It’s hard to believe that could be it.

@Bandersnatch I attached my raw data

#### Attached Files:

• ###### Newton.xlsx
File size:
47.8 KB
Views:
75
8. Jul 29, 2014

### Bandersnatch

Sorry, I can't open the file.

9. Jul 29, 2014

### Crovati

hmm weird. I put it in word instead

10. Jul 29, 2014

### Crovati

....

#### Attached Files:

• ###### NSL.docx
File size:
100.8 KB
Views:
125
11. Jul 29, 2014

### Bandersnatch

The data looks fine.

How was velocity measured? Are you sure you didn't put average velocity for each trial instead of final velocity?

12. Jul 29, 2014

### Crovati

Well for each trial we used a motion sensor and software program that took the velocity and time and graphed it. Then we chose an area from the graph and took ΔV/Δt to get the acceleration.