Kinematics: acceleration on a an incline (LAB)

In summary, the person was trying to find the value of g which should be around 9.8 m/s^2, but their answer was much higher due to incorrectly calculating the angle from the heights. They used the equation a=gsin(theta) to calculate the value of theta, and then used this to calculate g. They also used a graph to plot the value of d vs t^2 and found that their answer was 3.2 m/s^2. However, they realized their mistake and switched the values on the first graph. This allowed them to get the correct answer of 6.4 m/s^2.
  • #1
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Homework Statement


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.

Homework Equations


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

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 =)
 
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  • #2
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.
 
  • #3
BruceW said:
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.

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 traveled 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 traveling that fast
 
  • #5
Was the time displayed in seconds?
 
  • #6
grzz said:
Was the time displayed in seconds?

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 don't believe the angle is wrong but there might be an error with the length or timing
 
  • #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 !
 
  • #8
ha, no worries. Looks like you figured it out yourself in the end!
 

1. What is kinematics?

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion.

2. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is measured in meters per second squared (m/s²).

3. How is acceleration on an incline different from acceleration on a level surface?

Acceleration on an incline is affected by the angle of the incline and the force of gravity acting on the object. On a level surface, the only factor affecting acceleration is the force of gravity.

4. How do you calculate acceleration on an incline?

To calculate acceleration on an incline, you need to know the angle of the incline, the mass of the object, and the force of gravity. The formula is a = gsinθ, where a is acceleration, g is the force of gravity (9.8 m/s²), and θ is the angle of the incline.

5. What is the purpose of conducting a lab on acceleration on an incline?

The purpose of conducting a lab on acceleration on an incline is to observe and measure how the angle of an incline affects the acceleration of an object. This can help us better understand the principles of kinematics and how they apply to real-world situations.

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