# Homework Help: Theoretical Acceleration

1. Jul 30, 2009

### Petrucciowns

Hi, I am doing an acceleration/force lab, and I am having difficulty finding the theoretical acceleration.

In my lab it states:
The acceleration is the ratio of the net force divided by the total mass?

ath= (m-hanger*g-f) / m-cart+ m-hanger)

My values are:

M hanger= .00567 kg or 5.67 grams
M cart = .103 kg or 103 grams
Net force = .362

When I input the data into the equation I get:-2.82 kg m/s ^2, which first of all I have no clue how to get rid of the "kg" ( I know kg m/s ^2 = 1 N) , and secondly I know this is wrong ,because my actual value not theoretical came out to .332 m/s ^2.

Can anyone help me?

Thanks,

I have the complete document if you would like to take a look

2. Jul 30, 2009

### rock.freak667

$$a=\frac{Net \ Force}{m_{cart}+m_{hanger}}$$

Now if you put in those values you wrote down, I think you get 3.33 and not .333

the answer you will get will be m/s2.

Force is measured in Newtons(N). 1 N =1 kg m/s2

mass in kg.

$$a = \frac{N}{kg}= \frac{kg m/s^2}{kg} = m/s^2$$

3. Jul 30, 2009

### cepheid

Staff Emeritus
I don't know if the complete document is necessary, but a description of your experiment would certainly be useful!

I will say this. That equation should give you something in m/s^2. There is no way that kg should be left over, because you have something with units of mass in the numerator and something with units of mass in the denominator.

4. Jul 30, 2009

### Petrucciowns

Ok, my experiment consists of a physics cart. It has string tied to it with a paper clip on the end. The paper clips starts out with 1 quarter on it as a weight which is hung over a pulley tightened to the table. A motion sensor sensor is connected to my pc which measures distance on the y coordinate and time on the x coordinate. Each run I add a quarter and compare the acceleration. The thing is my first run was .333 m/s and the theoretical value was 3.33 so obviously I am doing something wrong? Maybe one of the previous calculations were wrong?

5. Jul 30, 2009

### cepheid

Staff Emeritus
rock.freak already explained in post#2 that your calculations were wrong adn teh theoretical value is in fact 3.33 m/s^2

6. Jul 30, 2009

### Petrucciowns

No you don't understand the .333 m/s was the value from the computer software picked up by the motion sensor.

7. Jul 30, 2009

### cepheid

Staff Emeritus
8. Jul 30, 2009

### Petrucciowns

That must be it. The calculation is shown for frictional force is:

f= F applied - Mt1*a1

Those values are: Mt1= .109 g a1=.332 m/s ^2 and I found F applied by m*g which were:
m= .109 kg (the cart and 1 quarter) and g= 9.80 m/s ^2 (acceleration due to gravity)

Also it says above in the document:
Run # 1 will be used to calculate the frictional force by finding the difference between the theoretical acceleration and the measured acceleration. The frictional force will then be subtracted from the applied force in runs #2-4 to find the net force.

9. Jul 30, 2009

### cepheid

Staff Emeritus
Here is the mistake. The total mass is the mass of the cart + the quarter. But the cart is not falling under the influence of gravity, is it? Therefore, the cart is not contributing to the applied force, and it makes no sense to multiply its mass by g. Do you understand?

10. Jul 30, 2009

### Petrucciowns

I see, well how do I find F applied then?

11. Jul 30, 2009

### cepheid

Staff Emeritus
Well, *something* is falling under the influence of gravity, causing the whole system to accelerate. The applied force is the weight of that something.

12. Jul 30, 2009

### Petrucciowns

Awesome thanks, I can't believe I didn't realize that. I guess I was just really stressed out.

Last edited: Jul 30, 2009