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#