Question

Чему равна сила упругости , с которой буксирный трос жесткостью 10^6 Н/м действует на автомобиль если при буксировке автомобиля трос удлиняется на 2 см.?

Answer 1

F=r*x; k=10^6; x=2*10^-2;

Answer 2

Calculation of the Elastic Force

To calculate the elastic force with which a tow rope of stiffness 10^6 N/m acts on a car when the rope elongates by 2 cm during towing, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring or elastic material is directly proportional to the displacement or elongation of the material.

The formula for Hooke's Law is:

F = k * x

Where: - F is the force exerted by the spring or elastic material, - k is the stiffness or spring constant, - x is the displacement or elongation of the material.

In this case, the stiffness of the tow rope is given as 10^6 N/m, and the elongation of the rope is 2 cm, which is equivalent to 0.02 m.

Substituting the given values into the formula, we can calculate the elastic force:

F = (10^6 N/m) * (0.02 m)

Calculating this expression gives us:

F = 20,000 N

Therefore, the elastic force with which the tow rope acts on the car is 20,000 N.