Question

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.

Answer 1

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`