Question

A car jack works when a mechanic uses a force 3.8 N, on a small piston of radius 10 cm. What force can he exert on a larger piston of radius 0.22 m?

Answer 1

18.392 N

Step-by-step explanation:

Given that,

Smaller force acting on the car, F₁ = 3.8 N

Radius of smaller radius, r₁ = 10 cm = 0.1 m

The radius of the larger piston, r₂ = 0.22 m

We need to find the force exerted on the larger piston. Let it is F₂.

Using the relation,

`(F_1)/(F_2)=(A_1)/(A_2)\\\\(F_1)/(F_2)=(\pi r_1^2)/(\pi r_2^2)\\\\(F_1)/(F_2)=(r_1^2)/(r_2^2)\\\\F_2=(F_1r_2^2)/(r_1^2)\\\\F_2=(3.8* (0.22)^2)/((0.1)^2)\\\\F_2=18.392\ N`

Hence, the force exerted on the larger piston is 18.392 N.