A hydraulic lift is used to jack a 920 kg car 11 cm off the floor. The diameter

Question

A hydraulic lift is used to jack a 920 kg car 11 cm off the floor. The diameter of the output piston is 18 cm, and the input force is 250 N.

(a) What is the area of the input piston?

______ m^2

(b) What is the work done in lifting the car 11 cm?

______ J

(c) If the input piston moves 13 cm in each stroke, how high does the car move up for each stroke?

______M

(d) How many strokes are required to jack the car up 11 cm? (Include fractions of a stroke in your answer).

_____

(e) Calculate the combined work input of all of the strokes.

______ J

Explanation / Answer

The diameter of the output piston is 18 cm, and the input force is 250 N.
Area of output = (9 cm)^2 = 254.47 cm^2 =0.0254 m^2

(a) What is the area of the input piston?
F/A (input) = F/A (output)
A (input) = F (input) *A (output) / F (output) = 250 N *0.025447 m^2 /( 920 kg *9.8 m/s^2) =
7.0561e-04 m^2

(b) What is the work done in lifting the car 11 cm?
920 kg *9.8 m/s^2 *0.11m =991.76 J

c) If the input piston moves 13 cm in each stroke, how high does the car move up for each stroke?
Work/stroke = 250 N * 0.13 m = 32.5 J
distance = Work/force = 32.5 J / (920 kg *9.8 m/s^2) = 3.60470275*10-3 m/stroke

d) 0.11 m / (3.60470275*10-3) = 30.5 strokes

e) 30.5 strokes * 32.5 J/ stroke = 991.76 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.