Finding Theoretical Acceleration in an Acceleration/Force Lab

AI Thread Summary
The discussion focuses on a physics lab involving acceleration and force calculations. The user struggles to find the theoretical acceleration, initially calculating it incorrectly as -2.82 kg m/s² instead of the expected 3.33 m/s². Key points include the correct understanding that net force should be measured in Newtons and that the total mass for calculations should only include the mass of the falling object, not the cart. The user learns that the applied force is determined by the weight of the object causing the acceleration, which resolves their confusion. Overall, the conversation emphasizes the importance of accurate calculations and understanding the principles of force and mass in physics experiments.
Petrucciowns
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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
 
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Your formula is basically

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.


so dimensionally your formula gives

a = \frac{N}{kg}= \frac{kg m/s^2}{kg} = m/s^2
 
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.
 
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?
 
Petrucciowns said:
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?

rock.freak already explained in post#2 that your calculations were wrong adn teh theoretical value is in fact 3.33 m/s^2
 
No you don't understand the .333 m/s was the value from the computer software picked up by the motion sensor.
 
That must be it. The calculation is shown for frictional force is:

f= F applied - Mt1*a1Those 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.
 
Petrucciowns said:
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)

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
I see, well how do I find F applied then?
 
  • #11
Petrucciowns said:
I see, well how do I find F applied then?

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