In preparing to shoot an arrow, an archer pulls a bowstring back 0.49 m by exerting a force that increases uniformly from 0 to 277 N. What is the equivalent spring constant of the bow?
average force=277N/2
277/2=k * .49
solve for k
To solve this problem, we need to understand the relationship between force, displacement, and spring constant. In this case, the bowstring is being pulled back, creating a displacement.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement it undergoes, and the constant of proportionality is known as the spring constant (k).
The equation that represents Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this problem, the force (F) exerted by the archer increases uniformly from 0 to 277 N, and the displacement (x) is 0.49 m.
We can use this information to find the spring constant (k).
Step 1: Calculate the average force.
Since the force increases uniformly, we can find the average force by taking the midpoint between 0 N and 277 N: (0 N + 277 N) / 2 = 138.5 N.
Step 2: Apply Hooke's Law to find the spring constant.
We know that F = -kx, so we can rearrange the equation to solve for k: k = -F / x.
Substituting the values we have, k = -138.5 N / 0.49 m.
Step 3: Calculate the spring constant.
Using a calculator, divide -138.5 N by 0.49 m to get the value of k.
The equivalent spring constant of the bow is approximately -282.653 N/m.