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Kinematics: acceleration on a an incline (LAB)

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Hey guys, I'm having a problem with my lab report write-up. The purpose of the lab is to find the value of g which should be around 9.8 m/s^2. This was found by releasing a cart from rest on an inclined plane and measuring the time it took to travel across a certain distance. I got the acceleration on the plane but my g is much higher than 9.8 which I believe is due to not correctly calculating the angle from the heights.

    2. Relevant equations
    d = 1/2*a*t^2
    a = gsin(theta)

    3. The attempt at a solution
    I have plotted d vs t^2 and got a value of 3.2 m/s^2.

    http://imageshack.us/photo/my-images/4/graph1p.jpg/

    using d = 1/2*a*t^2
    a = 2 (3.2 m/s^2)
    a = 6.4 m/s^2

    here is where i'm having the problem. We were told to take measurements of the height from where it started to where we stopped timing.

    http://imageshack.us/photo/my-images/522/graph2e.jpg/

    Using this I got:

    Sin(theta) = 13.04-9.2/69.47 = 0.055

    using a = gsin(theta)
    g = 6.4 m/s^2 / 0.055
    g = 115 m/s^s

    As you can see my g is way off from the theoretical value. I maybe doing something wrong. Any help would be much appreciated. Thanks for taking the time out to read it =)
     
  2. jcsd
  3. Nov 12, 2011 #2

    BruceW

    User Avatar
    Homework Helper

    Right. Your answer for 'a' seems OK (as long as d represents the distance it went along the slope, right?)

    And you've got the equation a=gsin(theta) so you also need the value of theta. I'm confused how you calculated theta. On the diagram, it looks like you measured the horizontal distance, but in the equation you seem to be using the vertical distance. If you used the horizontal distance, this would give you cos(theta), so this would affect the answer.

    Also, your answer for g is 11.5 m/s^2 And this isn't drastically far from the true answer. Maybe you could use several heights to calculate an average for theta. This might get a better answer.
     
  4. Nov 12, 2011 #3
    Hey thanks for the reply! The problem is actually that my answer is 115 METERS per second which is why I was saying it was way off the original number. It looks like there is a decimal error but I converted cm to meters correctly.

    For the theta (in the diagram) the 69.47 cm is the hypotenuse so the distance travelled by the cart and difference of the 2 heights is the vertical distance.

    Sin = opposite/hypotenuse = 13.04-9.2/69.47 = 0.055 is how I got the angle.

    I was wondering if my "a" is too big because it didn't seem like it was travelling that fast
     
  5. Nov 12, 2011 #4
  6. Nov 12, 2011 #5
    Was the time displayed in seconds?
     
  7. Nov 12, 2011 #6
    yes it was. There is a second part to the lab where the inital velocity is not equal to 0. I used the same angle of 0.055 and found the value of g as 9.45 m/s^2. So now I dont believe the angle is wrong but there might be an error with the length or timing
     
  8. Nov 12, 2011 #7
    I figured out my mistake, I switched the distance and time on the first graph.

    Thank you all for your help I really appreciate it !
     
  9. Nov 12, 2011 #8

    BruceW

    User Avatar
    Homework Helper

    ha, no worries. Looks like you figured it out yourself in the end!
     
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