(a) Find the force P that must be applied to a piston of area 12.10 cm 2 to prod

Question

(a) Find the force P that must be applied to a piston of area 12.10 cm2 to produce sufficient fluid pressure to support a car weighing 11,851 N by means of a column of fluid of cross sectional area 250 cm2 as seen in the figure below.


(b) Find the increase in the car's gravitational potential energy when it is raised 1.00 m.


(c) How far must the smaller piston move in order for the larger one to move 1.00 m?


(d) Calculate the work done by P in moving the smaller piston.


Compare your answer with the answer to part b.

Explanation / Answer

W = weight of the car = 11,851 N
Ap = cross-sectional area of the piston = 12.10 cm²
Ac = cross-sectional area of the column of fluid = 250 cm²

12.10/250 = P/11,851

250P = (12.10)(11,851)

250P = 143397.1
P = 143397.1/250

P = 573.58 N ANSWER

(b) Find the increase in the car's gravitational potential energy when it is raised 1.00 m.

PEg1 = gravitational potential energy = Wh
W = weight of the car = 11,851 N
h = height to which the car is raised = 1.00 m

PEg1 = Wh
PEg1 = (11851 N)(1.00 m)
PEg1 = 11851 Nm = 11851 J = 11851 kJ

PEg0 = gravitational potential energy at zero height = 0

So, the increase in the car's gravitational potential energy is:

PEg = PEg1 - PEg0 = 11851 Nm = 11851 J = 11851 kJ ANSWER

(c) How far must the smaller piston move in order for the larger one to move 1.00 m?

Ratio = 250/12.10 = 20.66

So,

d = distance traveled of smaller cylinder = to be determined

d = 22.66(1.00 m) = 22.66 m ANSWER

(d) Calculate the work done by P in moving the smaller piston.

W = work done = to be determined
P = applied force on the smaller piston = 573.58 N
d = distance traveled by smaller cylinder = 20.66 m

W = Pd
W = (572.58N)(20.66 m)
W = 11851 Nm = 11851 J = 11851 kJ ANSWER

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