Hydraulic patient lifting

A dentist lifted a patient in a chair by 10 cm using a hydraulic device. The mass of the patient with the chair is 100 kg.
a) What force did the dentist have to apply to the small piston, if the ratio of the areas of the large and small pistons is S2/S1=20?
b) How much work did the dentist perform when lifting the patient?
c) How many times did the dentist have to push the small piston, if one push covered a distance of 10 cm?

Final Answer:

F₁ =  50 N
W =  100 J
n =  20

Step-by-step explanation:

$$\begin{gathered} s_1=10\text{cm}\to \text{m}=10:100\text{m}=0.1\text{m} \\ m=100\text{kg} \\ g=10{\text{m/s}}^2 \\ k=20 \\ k=S_2/S_1 \\ {F}_2=m\cdot g=100\cdot 10=1000\text{N} \\ {p}_1=p_2 \\ {F}_1/S_2=F_2/S_1 \\ {F}_1/F_2=S_2/S_1 \\ {F}_1/F_2=1/k \\ {F}_1=F_2/k=1000/20=50\text{N} \end{gathered}$$

$$\begin{gathered} W=m\cdot g\cdot \Delta h \\ W=m\cdot g\cdot s_1=100\cdot 10\cdot 0.1=100\text{J} \end{gathered}$$

$$\begin{gathered} s_2=10\text{cm}\to \text{m}=10:100\text{m}=0.1\text{m} \\ {V}_1=V_2 \\ {S}_1\cdot s_1=S_2\cdot n\cdot s_2 \\ {S}_1\cdot s_1=S_2\cdot n\cdot s_2 \\ k\cdot s_1=n\cdot s_2 \\ n=k\cdot s_1/s_2=20\cdot 0.1/0.1=20 \end{gathered}$$

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