An arrow, starting from rest, leaves the bow with a speed of 16.5 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?
To find the speed at which the arrow would leave the bow if the average force exerted on it were doubled, we can use the principle of conservation of mechanical energy.
The initial kinetic energy of the arrow is given by the equation:
KE_initial = (1/2) * m * v_initial^2
where m is the mass of the arrow, and v_initial is the initial speed of the arrow.
We are assuming that all other factors remain the same except for the average force exerted on the arrow by the bow. If the force is doubled, then the work done on the arrow is also doubled.
The work done on the arrow is given by the equation:
W = F * d
where F is the average force applied on the arrow, and d is the distance over which the force is applied.
Since the work done is equal to the change in kinetic energy, we can write:
W = KE_final - KE_initial
Now, if the average force is doubled, the work done on the arrow (W) is also doubled. Therefore:
2W = KE_final - KE_initial
Since the arrow starts from rest, the initial kinetic energy (KE_initial) is zero. Therefore, the equation simplifies to:
2W = KE_final
Since the work done on the arrow is doubled, the final kinetic energy of the arrow (KE_final) is also doubled.
Therefore:
KE_final = 2 * (1/2) * m * v_final^2
Simplifying further:
KE_final = m * v_final^2
Since KE_final = 2W, we can substitute that into the equation:
m * v_final^2 = 2W
We can rearrange this equation to solve for the final speed (v_final):
v_final^2 = (2W) / m
Taking the square root of both sides, we get:
v_final = sqrt((2W) / m)
Therefore, the speed at which the arrow would leave the bow if the average force exerted on it were doubled can be calculated using the equation above, where W is the work done on the arrow and m is the mass of the arrow.