Answered

An ideal spring stretches 0.7 meters with a spring constant of [tex]k = 50 \, \text{N/m}[/tex]. What is the applied force?

Hint: See Section 10.1 in the textbook.



Answer :

To determine the applied force on a spring that is stretched, we can use Hooke's Law. Hooke's Law gives a relationship between the force exerted by a spring and its displacement from the equilibrium position. The law is represented by the formula:

[tex]\[ F = k \cdot x \][/tex]

where:
- [tex]\( F \)[/tex] is the force applied on the spring,
- [tex]\( k \)[/tex] is the spring constant, which measures the stiffness of the spring,
- [tex]\( x \)[/tex] is the displacement or stretch distance of the spring from its equilibrium position.

In this problem, we are given the following data:
- The spring constant [tex]\( k \)[/tex] is 50 N/m.
- The stretch distance [tex]\( x \)[/tex] is 0.7 meters.

We can substitute these values into the formula to find the applied force:

[tex]\[ F = 50 \, \text{N/m} \cdot 0.7 \, \text{m} \][/tex]

Multiplying these values:

[tex]\[ F = 35 \, \text{N} \][/tex]

Thus, the applied force on the spring is [tex]\( 35 \, \text{N} \)[/tex].