A person does 250 joules of work in pulling back the string of an archery bow. What is the initial speed of a 0.25-kilogram arrow the moment after it is fired from the bow?
To find the initial speed of the arrow, we need to apply the principle of conservation of mechanical energy. The work done in pulling back the string of the bow is equal to the change in potential energy, which is then converted into kinetic energy when the arrow is fired. We can use the following equation:
Work Done = Change in Potential Energy = Change in Kinetic Energy
The work done in pulling back the string is given as 250 joules. The potential energy is initially zero, as the string is in its relaxed position. However, immediately after firing, the arrow reaches its maximum height and the entire potential energy is converted to kinetic energy. Hence, the change in potential energy is equal to the final kinetic energy of the arrow.
The formula for kinetic energy is:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
Given that the mass of the arrow is 0.25 kilograms, we can substitute these values into the equation:
250 Joules = (1/2) * 0.25 kg * velocity^2
Now, let's solve for the velocity:
250 Joules = (1/2) * 0.25 kg * velocity^2
Divide both sides of the equation by 0.125 to isolate the velocity term:
2000 = velocity^2
Taking the square root of both sides:
√2000 = velocity
Therefore, the initial speed of the arrow, the moment after it is fired from the bow, is approximately 44.7 meters per second (m/s).
To find the initial speed of the arrow, we need to use the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.
The work done on the arrow is given as 250 joules. We can assume that all this work is converted into the kinetic energy of the arrow.
The kinetic energy formula is:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the arrow, and v is the velocity (initial speed) of the arrow.
Rearranging the formula, we have:
v^2 = (2KE) / m
Substituting the given values:
v^2 = (2 * 250 J) / 0.25 kg
v^2 = 1000 J / 0.25 kg
v^2 = 4000 J/kg
Taking the square root of both sides, we find:
v = √(4000 J/kg)
Calculating the square root of 4000 J/kg, we get:
v ≈ 63.25 m/s
Therefore, the initial speed of the arrow after being fired from the bow is approximately 63.25 m/s.