If a track is 1.7 m long and has a height change, Deltah, of 20 ± 0.2 cm, what is the acceleration of a frictionless cart sliding down the track, with uncertainty?
I'm unsure what to do here.
I'm unsure what to do here.
1 Answer
This is what I think is required to be done.
Explanation:
Given that a track is
Angle of incline
For the middle value
Let
Weight of cart
=mg
whereg is acceleration due to gravity.
Component of weight acting along the incline which produces sliding acceleration
Using Newton's Second Law of motion, acceleration produced by this component
Inserting calculated value of angle we get
a=g\ sin6.71^@=0.117g\ ms^-2 ......(1)
For uncertainty in angle on the positive side
theta=arctan(((20.2)/100)/1.7)=6.78^@
Corresponding acceleration
a_+=g\ sin6.78^@=0.118g\ ms^-2
Difference between the mean value acceleration
Deltaa_+=+0.001g\ ms^-2 .....(2)
Similarly, for uncertainty in angle on the negative side
theta=arctan(((19.8)/100)/1.7)=6.64^@
Corresponding acceleration
a_(-)=g\ sin6.64^@=0.116g\ ms^-2
Difference between the mean value acceleration
Deltaa_(-)=-0.001g\ ms^-2 .....(3)
Combining (1), (2) and (3)
acceleration
a=(0.117+-0.001)g\ ms^-2