In one stroke, the small piston of the hydraulic press descends to a distance of h1 = 0.2 m, and the large piston

In one stroke, the small piston of the hydraulic press descends to a distance of h1 = 0.2 m, and the large piston rises to a height of h2 = 0.01 m. With what force F2 does the press act on the body clamped in it, if the force F1 = 500N?

We divide one height by another: 0.1: 0.2 = 2. Therefore, the second piston is acted upon by a force twice as large as the first. Let’s find out what this force is equal to: 500 * 2 = 1000.
Answer: the piston acts on the body clamped by it with a force of 1000N.