A hydraulic lift is used to jack a 960-kg car 42 cm off the floor. The diameter

Question

A hydraulic lift is used to jack a 960-kg car 42 cm off the floor. The diameter of the output piston is 19 cm, and the input force is 350 N. What is the area of the input piston? Express your answer using two significant figures. A = _____ m^2 What is the work done in lifting the car 42 cm? Express your answer using two significant figures. W = ________ J If the input piston moves 12 cm in each stroke, how high does the car move up for each stroke? Express your answer using two significant figures. h = ______ How many strokes are required to jack the car up 42 cm? Express your answer using two significant figures.

Explanation / Answer

Given,

Mass, m = 960 kg

Diameter of piston, d = 19 cm

output force, F = mg = 960 x 9.8 = 9408 N

input force, f = 350 N

So, the ratio of forces is 9408 / 350 = 26.88

a) Area of the input = area of the output/26.88 = (pi x 0.0952)/26.88 = 0.001 m2

b) Work done, W = m x g x y = 960 x 9.8 x 0.42 = 3951.36 J

c)The volume of fluid moved is constant so (A x h)input = (A x h)output

If hinput = 12 cm,

then, houtput = (A x h)input / houtput = hinput/26.88 = 0.12/26.88 = 0.0045 m

d) 0.42 = n x 0.0045

=> n = 0.42/0.0045 = 94 strokes

Please rate my answer if you find it helpful, good luck....

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