# The box is pulled along the ground by a rope along a horizontal circle with a length of L = 40 m

**The box is pulled along the ground by a rope along a horizontal circle with a length of L = 40 m with a constant modulus speed. The work of the traction force for one revolution around the circumference A = 2.4 kJ. What is the modulus of the friction force acting on the box from the ground?**

Given:

L = 40 meters – the length of the circle along which the box is pulled;

A = 2.4 kJ = 2400 Joules – the work done by the pulling force in one revolution around the circumference.

It is required to determine Ftr (Newton) – the modulus of the friction force.

Let’s find the value of the traction force with which the box is pulled along the horizontal circle:

F = A / L = 2400/40 = 60 Newtons.

According to the condition of the problem, the box is pulled along the ground at a constant modulo speed, that is, evenly. Then, according to Newton’s first law:

F – Ftr = 0;

Ftr = F = 60 Newton.

Answer: the modulus of the friction force acting on the box is 60 Newtons