- #1
deanchhsw
- 14
- 0
Hello, I've just joined this forum in hopes of finding some help to my physics problems. I usually can find out what to do by myself, but this problem seems to be requiring some advanced concepts...or it maybe is so that I'm just not using my noodle.
A car is sliding down an inclined plane, say, of 20 degrees to the horizontal.
And the distance traveled, s, is 2.0818 meters, and the time elapsed is
2.33 seconds.
therefore, the calculated value of acceleration of the cart is .767m/s^2.
It is very easy up to this point, however, the teacher asks us to solve for acceleration due to gravity, otherwise known as 9.8 m/s^2, to compare the error margin to the accepted value. However, it doesn't come to me how you can solve for acceleration due to gravity just with the cart's acceleration.
Essentially, the only force of acceleration applied to the cart is that of the gravity, but the incline is reducing the accelertion, as it is not the "free fall".
which would be the case if the incline was 90(or 270) degrees. So my assupmtion is that there must be a certain ratio to solve this...but I just can't hold a grasp of it. I tried to get a brain blast(ha), didn't work for 20 minutes and I just ended up with a headache.
Of course, since one is solving for Ag, it cannot be incorporated in solving process- .. I have not been this frustrated with a problem for a long time..
Please help, and I will be greatful.
for your convenience, the data again is summed up here:
displacement = 2.0818 m
time = 2.33 s
acceleration cart = 0.767m/s^2
angle of the incline = 20
solve for Acc. due to gravity.
Once again, thank you, Dean.
* p.s. air resistance and friction is considered negligible in this lab.
A car is sliding down an inclined plane, say, of 20 degrees to the horizontal.
And the distance traveled, s, is 2.0818 meters, and the time elapsed is
2.33 seconds.
therefore, the calculated value of acceleration of the cart is .767m/s^2.
It is very easy up to this point, however, the teacher asks us to solve for acceleration due to gravity, otherwise known as 9.8 m/s^2, to compare the error margin to the accepted value. However, it doesn't come to me how you can solve for acceleration due to gravity just with the cart's acceleration.
Essentially, the only force of acceleration applied to the cart is that of the gravity, but the incline is reducing the accelertion, as it is not the "free fall".
which would be the case if the incline was 90(or 270) degrees. So my assupmtion is that there must be a certain ratio to solve this...but I just can't hold a grasp of it. I tried to get a brain blast(ha), didn't work for 20 minutes and I just ended up with a headache.
Of course, since one is solving for Ag, it cannot be incorporated in solving process- .. I have not been this frustrated with a problem for a long time..
Please help, and I will be greatful.
for your convenience, the data again is summed up here:
displacement = 2.0818 m
time = 2.33 s
acceleration cart = 0.767m/s^2
angle of the incline = 20
solve for Acc. due to gravity.
Once again, thank you, Dean.
* p.s. air resistance and friction is considered negligible in this lab.
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