A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a diameter of 0.15 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistons. A) 250 kg B) 700 kg C) 1400 kg D) 2800 kg E) 5600 kg

C. 1400

Explanation:

The force exerted = f1 = 250

Diameter d1 = 0.02

r1 = d1/2 = 0.01

Diameter d2 = 0.15

r2 = d2/2 = 0.075

The mass of jack to lift

F1/A1 = f2/A2

250/r1² = f2/r2²

250/0.01² = f2/0.075²

When we cross multiply, we will have:

250x0.005625 = 0.0001f2

1.40625 = 0.0001f2

F2 = 1.40625/0.0001

F2 = 14062.5N

Force f = mg

g = 9.81

m = 14062.5/9.81

Mass = 1433Kg

Therefore option c is correct. A mass of 1400kg can be lifted by the jack

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