Question

A construction system has a spring that must be extended 750 millimeters. If the spring’s k=100 N/m, Calculate the force that must be applied to the spring.

Answer 1

The force that is exerted on the spring is 75 N.

What is the force on the string?

Let us recall that Hooke's law states that the extension of an elastic material is directly proportional to the force that is applied as long as the elastic limit of the material has not been exceeded.

Given now the fact that we have;

F = Ke

F = force applied

K = force constant

e = extention

e = 750 millimeters or 0.75 m

K = 100 N/m

F =  0.75 m *  100 N/m

F = 75 N

Learn more about Hooke's law:brainly.com/question/13348278

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Answer 2

So, the force that must be applied to the spring is 75.0 N.

Introduction

Hi ! Here I will help you to solve the problem of the force required while stretching the spring. The force or load applied to the spring can cause the spring to stretch or compress. By stretching, there should be a change in the length of the spring ($\sf{\Delta x}$). The concept that you should know in this question :

• With the same spring constant, if the applied force is greater, then the value of the increase in length is also greater.
• With the same spring constant, if the applied force is greater, then the value of the increase in length is also greater.With the same applied force, if the value of spring constant is greater, the value of the increase in length is smaller.

Formula Used

The formula that show the relationship between the force, the coefficient of the spring, and the change in the length of the spring is expressed in the equation:

$\boxed{\sf{\bold{F = k \cdot \Delta x}}}$

With the following condition :

• F = the force that applied to spring (N)
• k = the spring constant (N/m)
• $\sf{\Delta x}$ = the change in the length of the spring (m)

Problem Solving

We know that :

• k = the spring constant = 100 N/m
• $\sf{\Delta x}$ = the change in the length of the spring = 750 mm = 0.75 m

What was asked ?

• F = the force that applied to spring = ... N

Step by step :

$\sf{F = k \cdot \Delta x}$

$\sf{F = 100 \cdot 0.75}$

$\boxed{\sf{F = 75.0 \: N}}$

Conclusion

So, the force that must be applied to the spring is 75.0 N.