Question

Груз массой 12 кг поднимают вверх с ускорением 2м/с2, прикрепив его к пружине жёсткостью 360H/м.На сколько удлинилась пружина?

Answer 1

R=ma (все векторно)
Fт+Fупр=ma
берем проекции на ось оу
oy: Fyпр-Fт=ma
kx-mg=ma
kx=m(a+g)
x=m (a+g)/k=12 (2+10)/360=12×12/360=0,4 (м)=40 (см)

Answer 2

Calculation of Spring Extension

To calculate the extension of the spring when lifting a 12 kg mass with an acceleration of 2 m/s^2 and attaching it to a spring with a stiffness of 360 N/m, we can use the formula for the force exerted by the spring.

The force exerted by the spring can be calculated using Hooke's Law:

F = kx

Where: - F is the force exerted by the spring - k is the stiffness of the spring - x is the displacement of the spring from its equilibrium position

The force exerted by the spring can also be calculated using the formula:

F = ma

Where: - F is the force - m is the mass - a is the acceleration

Equating the two formulas, we get:

kx = ma

Solving for x, we get:

x = ma / k

Calculation

Substituting the given values: - m = 12 kg - a = 2 m/s^2 - k = 360 N/m

We can calculate the extension of the spring:

x = (12 kg * 2 m/s^2) / 360 N/m

x = 0.0667 m

So, the spring extended by approximately 0.0667 meters when the 12 kg mass was lifted with an acceleration of 2 m/s^2 and attached to the spring with a stiffness of 360 N/m.