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.

1 Answer
Aug 11, 2018

This is what I think is required to be done.

Explanation:

Given that a track is 1.7\ m long and has a height change, Deltah= 20 ± 0.2\ cm.

Angle of incline theta=arctan(((20+-0.2)/100)/1.7)
For the middle value theta=arctan(((20)/100)/1.7)=6.71^@

Let m be mass of the friction less cart

Weight of cart =mg
where g is acceleration due to gravity.

Component of weight acting along the incline which produces sliding acceleration =mgsin theta
Using Newton's Second Law of motion, acceleration produced by this component a=(mgsin theta)/m=gsintheta

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