Question

A force of 200 N is applied to an input piston of cross-sectional area 2 sq. cm pushing it downward 2.8 cm. How far does the output piston of cross-sectional area 12 sq. cm move upward? Show all work.

Answer 1

Answer: 0.46 cm upward

Explanation:

Given that,

Input force
`F_(1) = 200 N`

Input piston of cross- sectional area
`A_(1) = 2\ cm^(2)`

Output force =
`F_(2)`

Output piston of cross-section area
`A_(2) = 12\ cm^(2)`

For output force,

Input pressure = Output pressure

`P_(1) = P_(2)`

`(F_(1))/(A_(1)) = (F_(2))/(A_(2))`

`F_(2) = (F_(1)* A_(2))/(A_(1))`

`F_(2) = (200\ N* 12\ cm^(2))/(2\ cm^(2))`

`F_(2) = 1200 N`

For upward displacement,

The work done

`W_(1) = W_(2)`

`F_(1)* d_(1) = F_(2)* d_(2)`

`d_(2) = (F_(1)* d_(1))/(F_(2))`

`d_(2) = (200\ N* 2.8\ cm)/(1200\ N)`

`d_(2) = 0.46\ cm`

Hence, the piston moves 0.46 cm upward.