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a spring is used to launch a ball vertically into the air. the spring has a spring constant of 200N/m and is compressed by 5 cm. if a ball of 10g is placed above the spring, what is the height reached by the ball assuming all the elastic potential energy is converted to gravitational potential energy?​

Respuesta :

MathPhys

Answer:

2.55 m

Explanation:

Elastic energy = gravitational energy

½ kx² = mgh

h = kx² / (2mg)

h = (200 N/m) (0.05 m)² / (2 × 0.010 kg × 9.8 m/s²)

h = 2.55 m

lhmarianateixeira

The height that the ball will assume is equal to 2.55 m

What is energy conservation?

This is called the Law of Conservation of Energy. In the case of a hydroelectric plant, for example, water flows in the river at a certain speed and falls from a certain height, turning like turbines, which transform mechanical energy into electrical energy.

So, this is Elastic energy = gravitational energy, making the calculus we have:

[tex](1/2)kx^2 = mgh\\h = kx^2 / (2mg)\\h = (200 N/m) (0.05 m)^2 / (2 * 0.010 kg * 9.8 m/s^2)\\h = 2.55 m[/tex]

See more about energy conservation at brainly.com/question/13949051

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