a spring of spring constant 5*10 3 N/m is stretched initially by 5cm

To understand the behavior of a spring when it is stretched, we can apply Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension or compression from its equilibrium position. The formula for Hooke's Law is given by:

Understanding Hooke's Law

The equation can be expressed as:

F = k * x

Where:

• F is the force exerted by the spring (in Newtons),
• k is the spring constant (in N/m), and
• x is the displacement from the equilibrium position (in meters).

Given Values

In your case, the spring constant (k) is 5,000 N/m, and the initial stretch (x) is 5 cm. To use the formula, we need to convert the displacement from centimeters to meters:

x = 5 cm = 0.05 m

Calculating the Force

Now, we can substitute the values into Hooke's Law:

F = 5,000 N/m * 0.05 m

Calculating this gives:

F = 250 N

Interpreting the Result

This means that when the spring is stretched by 5 cm, it exerts a force of 250 Newtons in the opposite direction, trying to return to its original length. This force is what you would feel if you were to pull on the spring or if it were attached to an object that was being pulled away from its equilibrium position.

Real-World Application

Understanding this principle is crucial in various fields, such as engineering and physics. For example, in designing suspension systems in vehicles, engineers must consider how springs behave under different loads to ensure safety and comfort.

Summary

In summary, by applying Hooke's Law, we determined that a spring with a constant of 5,000 N/m, when stretched by 5 cm, exerts a force of 250 N. This fundamental concept helps us grasp how springs function in both theoretical and practical scenarios.