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2 years ago

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An archer pulls her bowstring back 0.226 m by exerting a force that increases uniformly from zero to 263 N. What is the equivale

nt spring constant of the bow?

2 answers:

enyata [817]2 years ago

7 0

Answer: k = 1163.72 N/m²

Explanation: if the bow obeys hook's law, then force is proportional to extension.

F = ke.

F= Force = 263 N

k = force constant

e = extension = 0.226m

263 = k * 0.226

k = 263/ 0.226

k = 1163.72 N/m²

Georgia [21]2 years ago

6 0

<h2>Answer:</h2>

1163.7N/m

<h2>Explanation:</h2>

When a force caused by an object is applied to a spring, the spring applies an equal and opposite force to that object. This is evident in Hooke's law which states that the force(F) applied to a spring, either to cause extension or compression, is directly proportional to the length of the extension or compression (x). i.e

F = k x ---------------------(i)

Where k is the proportionality constant called <em>spring constant.</em>

<em />

<em>Given;</em>

F = force exerted on the bowstring = 263N

x = length of extension = 0.226m

<em>Substitute these values into equation (i) as follows;</em>

263 = k (0.226)

<em>Solve for k;</em>

k = 263 / 0.226

k = 1163.7N/m

Therefore, the spring constant of the bow is 1163.7N/m

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