# By applying a force of 300 N to the long end of the arm, the worker lifted a load of 1500 N to a height of 0.1 m.

**By applying a force of 300 N to the long end of the arm, the worker lifted a load of 1500 N to a height of 0.1 m. What work was done by the worker? How far was the end of the pole held by the worker? What work was done by gravity?**

F1 = 300 N.

P2 = 1500 N.

h2 = 0.1 m.

g = 9.8 m / s2.

A1 -?

h1 -?

A – ?

Simple mechanisms, such as the lever, do not give a gain in mechanical work, they give a gain in strength.

Work from opposite sides of the lever is equal to each other: A1 = A2.

A1 = F1 * h1.

A2 = P2 * h2.

F1 * h1 = P2 * h2.

h1 = P2 * h2 / F1.

h1 = 1500 N * 0.1 m / 300 N = 0.5 m.

A1 = 300 N * 0.5 m = 150 J.

We express the work of gravity A by the formula: A = m * g * h2 * cos180 °.

P2 = m * g.

A = P2 * h2 * cos180 °.

A = 1500 N * 0.1 m * (-1) = -150 J.

Answer: the worker has done work A1 = 150 J, the pole has dropped to a distance of h1 = 0.5 m, gravity is doing work A = -150 J.