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

$P_{1} = P_{2}$

$\dfrac{F_{1}}{A_{1}} = \dfrac{F_{2}}{A_{2}}$

$F_{2} = \dfrac{F_{1}\times A_{2}}{A_{1}}$

$F_{2} = \dfrac{200\ N\times 12\ cm^{2}}{2\ cm^{2}}$

$F_{2} = 1200 N$

For upward displacement,

The work done

$W_{1} = W_{2}$

$F_{1}\times d_{1} = F_{2}\times d_{2}$

$d_{2} = \dfrac{F_{1}\times d_{1}}{F_{2}}$

$d_{2} = \dfrac{200\ N\times 2.8\ cm}{1200\ N}$

$d_{2} = 0.46\ cm$

Hence, the piston moves 0.46 cm upward.