Answer
(a) A force of $625~N$ must be applied to the small piston.
(b) The car is lifted 6.25 mm
(c) The mechanical advantage is 16
Work Step by Step
(a) We can use Pascal's principle to find $F_a$:
$\frac{F_a}{a} = \frac{F_A}{A}$
$\frac{F_a}{\pi~r_a^2} = \frac{F_A}{\pi~r_A^2}$
$F_a = \frac{r_a^2~F_A}{r_A^2}$
$F_a = \frac{(2.50~cm)^2~(10,000~N)}{(10.0~cm)^2}$
$F_a = 625~N$
A force of $625~N$ must be applied to the small piston.
(b) The work done on the small piston is equal to the work done by the large piston on the car:
$F_A~d_A = F_a~d_a$
$d_A = \frac{F_a~d_a}{F_A}$
$d_A = \frac{(625~N)(10.0~cm)}{10,000~N}$
$d_A = 0.625~cm = 6.25~mm$
The car is lifted 6.25 mm
(c) We can find the mechanical advantage:
$\frac{W}{F_a} = \frac{10000~N}{625~N} = 16$
The mechanical advantage is 16.
