Finding spring constant k based on diagram -- Don't understand how to interpret force as drawn

[cestlavie]

Homework Statement

What is the spring constant k for the spring shown below? (Diagram attached below, and answers are in N/cm)

Relevant Equations

$F = kx$

$$F=kx$$
$$k=\frac F x= \frac {50+50~N} {5+5~ cm}= \frac {100~N} {10~cm}= 10~N/{cm}$$
However, the answer is $5~N/cm$, because the force on the spring is $50~N$. I am having trouble understanding why the force isn't $50~N$ + $50~N$. The diagram looks as though the spring is experiencing force on both sides, so shouldn't it be the sum of both? The only explanation I am given is to imagine the spring is positioned vertically and a $50~N$ weight is applied to it. Therefore, the floor will apply $50~N$ upward, making $F=50~N$. But I still see this scenario as having $100~N$ of force. Why isn't the weight accounted for in the equation?

 

[erobz]

Homework Statement: What is the spring constant k for the spring shown below? (Diagram attached below, and answers are in N/cm)
Relevant Equations: $F = kx$

View attachment 342440
$$F=kx$$
$$k=\frac F x= \frac {50+50~N} {5+5~ cm}= \frac {100~N} {10~cm}= 10~N/{cm}$$
However, the answer is $5~N/cm$, because the force on the spring is $50~N$. I am having trouble understanding why the force isn't $50~N$ + $50~N$. The diagram looks as though the spring is experiencing force on both sides, so shouldn't it be the sum of both? The only explanation I am given is to imagine the spring is positioned vertically and a $50~N$ weight is applied to it. Therefore, the floor will apply $50~N$ upward, making $F=50~N$. But I still see this scenario as having $100~N$ of force. Why isn't the weight accounted for in the equation?

Remove one of the 50 N forces, replace it with a wall. What is the force that the wall applies to the spring when the other end has 50 N applied?

 

[PeroK]

Homework Statement: What is the spring constant k for the spring shown below? (Diagram attached below, and answers are in N/cm)
Relevant Equations: $F = kx$

View attachment 342440
$$F=kx$$
$$k=\frac F x= \frac {50+50~N} {5+5~ cm}= \frac {100~N} {10~cm}= 10~N/{cm}$$
However, the answer is $5~N/cm$, because the force on the spring is $50~N$. I am having trouble understanding why the force isn't $50~N$ + $50~N$. The diagram looks as though the spring is experiencing force on both sides, so shouldn't it be the sum of both? The only explanation I am given is to imagine the spring is positioned vertically and a $50~N$ weight is applied to it. Therefore, the floor will apply $50~N$ upward, making $F=50~N$. But I still see this scenario as having $100~N$ of force. Why isn't the weight accounted for in the equation?

This is something many students get wrong when they first see it. If you apply a single $50N$ force to a spring, then (by Newton's second law), the spring accelerates in the direction of the force. In order to compress a spring, you must apply an equal force to both ends. This is how the spring constant is defined and measured:

If you apply a force of $50N$ to both ends of a spring and it compresses by a total of $10 cm$, then the spring constant is $500N/m$.

Note that, as @erobz points out, the second force may be explicitly specified, or it may be implied by a support at one end of the spring and the application of Newton's third law.

The same definition applies to tension in an extended elastic string. You must apply an equal force at each end of the string to extend it. If there were only one force pulling the string at one end, then the string would simply accelerate away in that direction. And, again, if the tension in a string is $10N$, say, then the string is being stretched by $10N$ is each direction - and itself providing an elastic force of $10N$ in both directions. And, again, that is the definition of tension.

 

[kuruman]

Why isn't the weight accounted for in the equation?

What weight?

One tends to think that when a spring is compressed such that one end is prevented from moving as in up against a wall only the end that moves can exert a force as if it is moving the end that creates the force. That is not the case. As soon as the distance between the ends is changed by $x$ from the relaxed length, equal and opposite forces of magnitude $kx$ appear at both ends of the spring. So here is another way of looking at it although @erobz's explanation is fine.

The overall displacement of the spring from its relaxed length is $x=10~$cm. This means that it exerts forces of magnitude $kx$ in opposite directions at each end. Since the spring does not accelerate, the net force on it must be zero. This means that $kx=50~$N. What do you get when you plug in $x=10!$ cm?

 

[haruspex]

You have to start with the definition of spring constant: if the spring length is reduced by x then the force with which it pushes at the end is increased by kx. But if it is pushing with kx at one end then, ignoring other forces such as its own weight or inertial force, it must be pushing with the same force at the other end.

 

[cestlavie]

In order to compress a spring, you must apply an equal force to both ends. This is how the spring constant is defined and measured:

As soon as the distance between the ends is changed by from the relaxed length, equal and opposite forces of magnitude appear at both ends of the spring.

Okay, this is what I was missing! So the second force in my and @erobz's example is Newton's third law in application. But because elastic force is applied on both ends, it would be incorrect to add $50~N$ twice as it's the same force. Thank you, everyone, for responding!