# A block with mass of 5.0kg is suspended from an ideal spring having negligible mass and stretches...

## Question:

A block with mass of 5.0kg is suspended from an ideal spring having negligible mass and stretches the spring 0.20m to its equilibrium position.

A) What is the force constant of the spring?

B) The spring is then stretched 0.80m from its equilibrium position and then released with the velocity zero. Calculate the velocity of the mass when the mass passes again through the equilibrium position.

## The Spring And Its Restoring Force:

When a spring of force constant {eq}\displaystyle { k } {/eq} is stretched or compressed from its equilibrium length say by {eq}\displaystyle { x } {/eq}, then a restoring force {eq}\displaystyle {F_{res}=-k x } {/eq} will act trying to restore it back to its initial shape. We can see that the restoring force is proportional to the displacement from the equilibrium position and always opposes the displacement.

## Answer and Explanation:

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A block of mass, {eq}\displaystyle {m=5.0 \ kg } {/eq} is suspended from a spring.

The block experiences a downward pull due to the gravitational...

See full answer below.

#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
202K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.