Final answer:
The elastic potential energy stored in the bow is 49 J. Only 50% of this energy, or 24.5 J, is transferred to the arrow. The velocity of the arrow depends on the mass of the arrow.
Step-by-step explanation:
To determine the amount of elastic potential energy stored in the bow, we can use the formula:
Elastic Potential Energy = (1/2)kx^2
where k is the spring constant of the bow and x is the displacement of the string from its equilibrium position.
In this case, the force required to hold the string at a displacement of 70 cm is 140 N. We can use this force to calculate the spring constant:
F = k * x
140 N = k * 0.7 m
Solving for k, we find that the spring constant is 200 N/m.
Now, we can calculate the elastic potential energy using the equation:
Elastic Potential Energy = (1/2) * 200 N/m * (0.7 m)^2
Calculating this expression, we find that the elastic potential energy stored in the bow is 49 J.
When the bow is released, only 50% of the elastic potential energy is transferred to the arrow. Therefore, the amount of energy transferred to the arrow is 0.5 * 49 J = 24.5 J.
Using the principle of conservation of energy, we can equate the elastic potential energy transferred to the arrow to the kinetic energy of the arrow:
24.5 J = (1/2)mv^2
where m is the mass of the arrow and v is its velocity.
Since the mass of the arrow is not provided, we cannot determine the exact velocity. However, we can rearrange the equation to solve for v:
v = sqrt((2 * 24.5 J) / m)
Inserting the mass of the arrow, we can calculate the velocity.