A 0.069-kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.95 m. With what speed does the arrow leave the bow?
Work done on arrow = Kinetic energy
65*0.95 = (M/2)V^2 = 0.0345 V^2
Solve for V, in m/s.
I get 42.3 m/s
To find the speed at which the arrow leaves the bow, we can use the principle of conservation of energy. The work done by the bowstring in accelerating the arrow is equal to the change in kinetic energy of the arrow.
The work done by the bowstring can be calculated using the formula:
Work = Force * Distance * cosθ
In this case, the force applied by the bowstring is 65 N, the distance over which the force is applied is 0.95 m, and the angle θ is 0 degrees (since the force is applied horizontally). As a result, the angle θ does not affect the calculation, since the cosine of 0 degrees is equal to 1.
So, the work done by the bowstring is:
Work = 65 N * 0.95 m * 1
Next, we can equate this work done with the change in kinetic energy of the arrow:
Work = Change in Kinetic Energy
The change in kinetic energy of the arrow can be expressed as:
Change in Kinetic Energy = (1/2) * mass * (final velocity)^2 - (1/2) * mass * (initial velocity)^2
Since the arrow is initially at rest, its initial velocity is 0.
Substituting the given values, the equation becomes:
65 N * 0.95 m = (1/2) * 0.069 kg * (final velocity)^2 - (1/2) * 0.069 kg * 0^2
Simplifying the equation, we have:
62.675 Nm = (1/2) * 0.069 kg * (final velocity)^2
We can now solve for the final velocity:
(final velocity)^2 = (2 * 62.675 Nm) / 0.069 kg
(final velocity)^2 = 1816.65 m^2/s^2
Taking the square root of both sides, we get:
final velocity = √(1816.65 m^2/s^2)
final velocity ≈ 42.67 m/s
Therefore, the arrow leaves the bow with a speed of approximately 42.67 m/s.