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.