(a) [tex]5.45\cdot 10^{-4} m^2[/tex]
According to Pascal's principle, the pressure on the first piston is equal to the pressure on the second piston:
[tex]p_1 = p_2\\\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex] (1)
where
F1 = 250 N is the input force
A1 = ? is the area of the input piston
F2 is the output force
A2 is the area of the output piston
The output force is just the weight of the car:
[tex]F_2 = mg =(1170 kg)(9.8 m/s^2)=11,466 N[/tex]
The radius of the output piston is half the diameter: [tex]r=d/2=18 cm/2 = 9 cm =0.09 m[/tex], so its area is
[tex]A_2 = \pi r^2 = \pi (0.09 m)^2=0.025 m^2[/tex]
So we can solve eq.(1) for A1, the area of the first piston:
[tex]A_1 = A_2 \frac{F_1}{F_2}=(0.025 m^2)\frac{250 N}{11,466 N}=5.45\cdot 10^{-4} m^2[/tex]
(b) 1491 J
The work done in lifting the car 13 cm is equal to the gravitational potential energy gained by the car:
[tex]W=\Delta U=mg \Delta h[/tex]
where:
m = 1170 kg is the mass of the car
g = 9.8 m/s^2
[tex]\Delta h=13 cm=0.13 m[/tex] is the increase in height of the car
Substituting,
[tex]W=\Delta U=(1170 kg)(9.8 m/s^2)(0.13 m)=1491 J[/tex]
(c) 0.0028 m
Assuming the machine is 100% efficient and there is no waste of energy, the input work is equal to the output work:
[tex]W_i = W_o\\F_1 d_1 = F_2 d_2[/tex]
where
F1 = 250 N is the input force
d1 = 13 cm = 0.13 m is the displacement of the input piston
F2 = 11,466 N is the output force (the weight of the car)
d2 is the displacement of the output piston
Solving for d2,
[tex]d_2 =d_1 \frac{F_1}{F_2}=(0.13 m)\frac{250 N}{11466 N}=0.0028 m[/tex]
(d) 46 strokes
In order to lift the car up 13 cm (0.13 m), we have to divide this value by the displacement of the car for each stroke, so we have:
[tex]n=\frac{0.13 m}{0.0028 m}=46.4 \sim 46[/tex]
(e) 1491 J
The work done during all of the strokes is equal to the gravitational potential energy gained by the car while being lifted 13 cm, so it is equal to the value found in part b):
W = 1491 J
