An archer pulls her bowstring back 0.488 m by extending a force that increases uniformly from zero to 251 N.
What is the equivalent spring constant of the bow?
Answer in units of N/m.
How much work does the archer do in pulling the bow?
Answer in units of J.
F = kx
251 = k (.488)
(1/2) k x^2 Joules
To find the equivalent spring constant of the bow, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Let's break down the problem step by step:
1. Determine the displacement of the bowstring:
The archer pulls the bowstring back a distance of 0.488 m.
2. Calculate the force:
The force increases uniformly from zero to 251 N. This means that at the maximum displacement, the force is at its peak value of 251 N.
3. Apply Hooke's Law:
Hooke's Law can be written as F = kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the force (251 N) corresponds to the maximum displacement (0.488 m). Substitute these values:
251 N = k * 0.488 m
4. Solve for the spring constant:
Divide both sides of the equation by 0.488 m to isolate k:
k = 251 N / 0.488 m
Now we can calculate the spring constant:
k ≈ 514.34 N/m
To find the work done by the archer, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
In this case, the bowstring starts from rest and ends at rest, so there is no change in kinetic energy. Therefore, the work done by the archer is equal to the potential energy stored in the bowstring.
The potential energy stored in a spring is given by the formula U = (1/2) * k * x^2, where U is the potential energy, k is the spring constant, and x is the displacement.
Substituting the values:
U = (1/2) * 514.34 N/m * (0.488 m)^2
Calculating the work done:
U ≈ 0.0599 J
Therefore, the archer does approximately 0.0599 J of work in pulling the bow.