Answer: 0.46 cm upward
Explanation:
Given that,
Input force [tex]F_{1} = 200 N[/tex]
Input piston of cross- sectional area [tex]A_{1} = 2\ cm^{2}[/tex]
Output force = [tex]F_{2}[/tex]
Output piston of cross-section area[tex]A_{2} = 12\ cm^{2}[/tex]
For output force,
Input pressure = Output pressure
[tex]P_{1} = P_{2}[/tex]
[tex]\dfrac{F_{1}}{A_{1}} = \dfrac{F_{2}}{A_{2}}[/tex]
[tex]F_{2} = \dfrac{F_{1}\times A_{2}}{A_{1}}[/tex]
[tex]F_{2} = \dfrac{200\ N\times 12\ cm^{2}}{2\ cm^{2}}[/tex]
[tex]F_{2} = 1200 N[/tex]
For upward displacement,
The work done
[tex]W_{1} = W_{2}[/tex]
[tex]F_{1}\times d_{1} = F_{2}\times d_{2}[/tex]
[tex]d_{2} = \dfrac{F_{1}\times d_{1}}{F_{2}}[/tex]
[tex]d_{2} = \dfrac{200\ N\times 2.8\ cm}{1200\ N}[/tex]
[tex]d_{2} = 0.46\ cm[/tex]
Hence, the piston moves 0.46 cm upward.
