Question

На скільки видовжиться пружина жорсткістю 400 H/m. Якщо до неї підвісили тіло масою 2 кг?

Answer 1

mg = kΔx
m = 2 кг
k = 400 H/м
g = 10 м с⁻¹
Δx = mg/k = 2·10/400 = 0.05 м (5 cм)

Answer 2

Calculation of Spring Extension

To calculate the extension of a spring with a given stiffness and a suspended mass, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. The formula for Hooke's Law is:

F = kx

Where: - F is the force exerted by the spring (in newtons) - k is the spring constant or stiffness (in newtons per meter) - x is the extension of the spring from its equilibrium position (in meters)

We can rearrange this formula to solve for the extension x:

x = F / k

Given that the stiffness of the spring is 400 H/m and the mass suspended from it is 2 kg, we can calculate the extension of the spring.

Calculation

First, we need to find the force exerted by the suspended mass. The force due to gravity can be calculated using the formula:

F = mg

Where: - F is the force due to gravity (in newtons) - m is the mass (in kilograms) - g is the acceleration due to gravity (approximately 9.81 m/s^2)

Substituting the given values: F = 2 kg * 9.81 m/s^2

F = 19.62 N

Now, we can use Hooke's Law to find the extension of the spring:

x = F / k

x = 19.62 N / 400 N/m

x ≈ 0.04905 meters

Conclusion

The spring would extend by approximately 0.04905 meters when a 2 kg mass is suspended from it.