Question

.На пружине жёсткостью 40 Н/м висит груз массой 200 грамм. Определите длину пружины. В недеформированном состоянии длина пружины равна 10 см.

Answer 1

) Новая длина : первоначальная + та часть, на которую растянулась пружина под действием груза. В растянутом состоянии компенсируются две силы - сила тяжести и сила упругости пружины (вспоминаем закон Гука):

 

 см

Длина пружины: 10+5=15 см

Answer 2

Calculation of the Length of the Spring

To determine the length of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is:

F = kx

Where: - F is the force applied to the spring, - k is the spring constant (also known as the stiffness constant), - x is the displacement of the spring from its equilibrium position.

In this case, we are given that the spring has a stiffness constant of 40 N/m and a mass of 200 grams (0.2 kg) is hanging from it.

To find the displacement of the spring, we can use the formula for weight:

F = mg

Where: - F is the force (weight) applied to the spring, - m is the mass of the object, - g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values into the equation, we have:

F = (0.2 kg) * (9.8 m/s^2)

Now, we can equate the force applied to the spring (F) with the force calculated using Hooke's Law (kx):

(0.2 kg) * (9.8 m/s^2) = (40 N/m) * x

Simplifying the equation, we find:

x = (0.2 kg * 9.8 m/s^2) / (40 N/m)

Now we can calculate the displacement (x) of the spring.