Under the action of mutually perpendicular forces 3Н and 4Н, the body moves in the direction of the resultant of these

Under the action of mutually perpendicular forces 3Н and 4Н, the body moves in the direction of the resultant of these forces at a distance of 15 m. What is the work of the resultant force?

Given:

F1 = 3 Newton – the first force;

F2 = 4 Newton – second force;

L = 15 meters – the distance the body moved under the action of the forces F1 and F2.

It is required to determine A (Joule) – the work of the resultant forces.

Since, according to the condition of the problem, the forces F1 and F2 are directed perpendicular to each other, we find the resultant of these forces:

F = (F1 ^ 2 + F2 ^ 2) ^ 0.5 = (3 ^ 2 + 4 ^ 2) ^ 0.5 = (9 + 16) ^ 0.5 = 25 ^ 0.5 = 5 Newtons.

Then the work of force will be equal to:

A = F * L = 5 * 15 = 75 Joules.

Answer: the work of the resultant force is 75 Joules.