To find the magnitude of the force stored in a spring, we can use Hooke's Law:

$F = k \times x$

where:

• $F$ is the force exerted by the spring (in newtons),
• $k$ is the spring constant (in newtons per meter),
• $x$ is the displacement or stretch of the spring (in meters).

Given:

• $k = 15 \, \text{N/m}$,
• $x = 50 \, \text{cm} = 0.50 \, \text{m}$ (converted to meters).

Now, we can substitute the values into the formula:

$F = 15 \, \text{N/m} \times 0.50 \, \text{m}$

$F = 7.5 \, \text{N}$

Result:

The magnitude of the force stored in the spring is 7.5 N.

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Would you like further details on this or have any questions? Here's a few related concepts to consider:

1. How does Hooke's Law relate to elastic potential energy in the spring?
2. What happens to the force if the spring is stretched further?
3. How does the spring constant affect the stiffness of the spring?
4. What is the difference between elastic and plastic deformation in springs?
5. How would the force change if the spring were compressed instead of stretched?

Tip: Always remember to convert your measurements to SI units before performing calculations!