A 0.077-kg arrow is fired horizontally. The bowstring exerts an average force of 50 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?
The work done by the bow string, F * X = 50 x 0.75 Joules,
equals the kinetic energy (MV^2/2) of the arrow as it leaves the bow.
Solve for V
To find the speed at which the arrow leaves the bow, we can use the principle of work and energy. The work done on the arrow by the bowstring is equal to the change in its kinetic energy.
The work done on the arrow can be calculated using the formula:
Work (W) = Force (F) * Distance (d)
In this case, the force exerted by the bowstring is 50 N, and the distance over which the force is applied is 0.75 m. Plugging in these values, we get:
W = 50 N * 0.75 m
W = 37.5 J
The work done is equal to the change in kinetic energy, so we can write:
W = ΔKE
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * mass * velocity^2
In this case, the mass of the arrow is 0.077 kg, and we need to find the velocity (v). Rearranging the formula, we get:
v = sqrt(2 * KE / mass)
Plugging in the values of the work done and the mass, we have:
v = sqrt(2 * 37.5 J / 0.077 kg)
v = sqrt(974.03 m^2/s^2)
v ≈ 31.2 m/s
Therefore, the arrow leaves the bow with a speed of approximately 31.2 m/s.