Cart down a ramp, acceleration of 0.6m/s^2

jnimagine
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We rolled down a cart down a ramp with a tickertape. When the average velocities were calulated and they were used to calculate acceleration. But we got acceleration of like 0.6, 0.2m/s^2 etc... shouldn't it be 9.8m/s^2..?! When i graphed the average velocity and the line of best fit was drawn and the slope calculated i got 6.5m/s^2. Why did i get an acceleration of such numbers like 0.6?? What are some explanations to account for this acceleration?
 
Last edited:
jnimagine said:
We rolled down a cart down a ramp with a tickertape. When the average velocities were calulated and they were used to calculate acceleration. But we got acceleration of like 0.6, 0.2m/s^2 etc... shouldn't it be 9.8m/s^2..?! When i graphed the average velocity and the line of best fit was drawn and the slope calculated i got 6.5m/s^2. Why did i get an acceleration of such numbers like 0.6?? What are some explanations to account for this acceleration?
In which direction does the cart's weight act?
 
Hootenanny said:
In which direction does the cart's weight act?

um.. the cart's weight acts down??
 
jnimagine said:
um.. the cart's weight acts down??
Down the ramp or directly down towards the centre of the earth?
 
Hootenanny said:
Down the ramp or directly down towards the centre of the earth?

down toward the earth... so there are two directions involved in this motion??
so 9.8m/s^2 is the acceleration down towards the earth... then what would be 0.6m/s^2??
 
cart down a ramp accelerates at 0.6m/s^2?? not 9.8m/s^2?

We rolled down a cart down a ramp with a tickertape. When the average velocities were calulated and they were used to calculate acceleration. But we got acceleration of like 0.6, 0.2m/s^2 etc... shouldn't it be 9.8m/s^2..?! When i graphed the average velocity and the line of best fit was drawn and the slope calculated i got 6.5m/s^2. Why did i get an acceleration of such numbers like 0.6?? What are some explanations to account for this acceleration?
 
The acceleration depends on the slope of the ramp. 9.8 m/s^2 would only be expected if the ramp were vertical--the cart was in free fall. :wink:
 
This is easy enough to calculate by finding the component of the gravitational force that is applied along the slope of the ramp. It's simple trigonometry.
 
Doc Al said:
The acceleration depends on the slope of the ramp. 9.8 m/s^2 would only be expected if the ramp were vertical--the cart was in free fall. :wink:

then what is 0.6m/s^2?
 
  • #10
jnimagine said:
then what is 0.6m/s^2?
Re-read Doc Al's post.
 
  • #11
Acceleration is equal to gravity only if the object is in free-fall. If the object is rolling or sliding down an inclined plane, the acceleration is dependent on friction and the angle of the incline. Remember that acceleration down an inclined plane is independent of the mass. Hope this helps!
 

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