Answer: An archer pulls her bowstring back 0.400m by exerting a force that increases uniformly from zero to 230N. The equivalent spring constant of the bow will be 575N/m
Explanation: To find the answer, we have to know about the Simple Harmonic Motion.
What is Simple Harmonic Motion?
- A particle is said to execute simple harmonic motion, if it moves to and fro about mean position under the action of restoring force.
- The restoring force is directly proportional to its displacement from the mean position and always directed towards the mean position.
- If x is the displacement from the mean position, and F is the restoring force, then
[tex]F[/tex]∝[tex]-x[/tex]
[tex]F=-kx[/tex] where, k is called the spring constant.
How to approach the problem?
- Given that, an archer pulls her bowstring back 0.400m by exerting a force that increasing from zero to 230N.
- Here, from the question given, we can write, [tex]x=-0.400m[/tex] and [tex]F=230N[/tex].
- Thus, our spring constant k will be,
[tex]k=-F/x= (230)/ (0.400) =575 N/m.[/tex]
Thus, we can conclude that, the equivalent spring constant of the bow will be 575N/m.
Learn more about the Simple Harmonic Motion here:
https://brainly.com/question/28019840
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