An archer pulls her bowstring back 0.226 m by exerting a force that increases uniformly from zero to 263 N. What is the equivalent spring constant of the bow?

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

1163.7N/m

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 spring constant.

Given;

F = force exerted on the bowstring = 263N

x = length of extension = 0.226m

Substitute these values into equation (i) as follows;

263 = k (0.226)

Solve for k;

k = 263 / 0.226

k = 1163.7N/m

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

f is prop to m1xm2 double m1 double force

ok

explanation: