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A hydraulic lift is designed to lift cars up to 2000 kg in mass.If the lift under the car is 1.0 m by 1.2m and the area of the input piston is 10 cm by 10 cm, how much force must be exerted on the input piston to lift a 2000-kg car? A) 136 N B 24x 10^4 N C) 160 N D) 196 N E) 17N

Respuesta :

Answer:

C) 160 N

Explanation:

[tex]m[/tex] = mass of the car = 2000 kg

Weight of the car is given as

[tex]W = mg \\W = (2000) (9.8)\\W = 19600 N[/tex]

[tex]F_{i}[/tex] = Force exerted on the input piston

[tex]A_{i}[/tex] = Area of input piston = 10 cm x 10 cm = 0.1 m x 0.1 m = 0.01 m²

[tex]A_{lift}[/tex] = Area of lift = 1 m x 1.2 m = 1.2 m²

Using Pascal's law

[tex]\frac{F_{i}}{A_{i}} = \frac{W}{A_{lift}}\\\frac{F_{i}}{0.01} = \frac{19600}{1.2}\\F_{i} = 160 N[/tex]

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