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2 years ago

8. A spring of spring constant 4 N/m is stretched 0.5 meters. How strong is the restoring force?

Physics

1 answer:

pogonyaev2 years ago

4 0

So, the recovery force of the spring is <u>2 N in the opposite direction of the pull</u>.

<h3>Introduction</h3>

Hi ! Here, I will help you about the spring recovery force. <u>The restoring force is the force that opposes the direction of the initial pull of the spring (either when the spring is pulled horizontally or vertically)</u>. The restoring force is strongly influenced by the type of spring (through the spring constant) and the length of the strain that occurs. Negative values in spring restoring force only indicate direction, not value. The equation that applies is as follows:

If the spring is pulled horizontally

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

With the following condition :

F = recovery force (N)
k = spring constant (N/m)
$\sf{\Delta x}$ = horizontal length change (m)

If the spring is pulled vertically

$\boxed{\sf{\bold{F = - k \times \Delta y}}}$

With the following condition :

F = recovery force (N)
k = spring constant (N/m)
$\sf{\Delta y}$ = vertical length change (m)

<h3>Problem Solving</h3>

We know that :

Assume the spring is pulled horizontally
k = spring constant = 4 N/m
$\sf{\Delta x}$ = horizontal length change = 0.5 m

What was asked :

F = recovery force = ... N

Step by step :

$\sf{F = - k \times \Delta x}$

$\sf{F = - 4 \times 0.5}$

$\boxed{\sf{F = -2 \: N}}$

<h3>Conclusion :</h3>

So, the recovery force of the spring is 2 N in the opposite direction of the pull.

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