An 0.075 kg arrow is pulled back in a bow. After being held still, it is let go, moving at 23.8 m/s after the 0.55 meters in which the arrow is in contanct with the bow string. What work was done on the arrow by the bow string? What force was exerted by the archer to pull the bowstring back?
To determine the work done on the arrow by the bow string, we can use the work-energy principle. The principle states that the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the arrow is 0 since it is held still, and the final kinetic energy is given by:
KE = (1/2) * m * v^2
Where:
m = mass of the arrow = 0.075 kg
v = final velocity of the arrow = 23.8 m/s
KE = (1/2) * 0.075 kg * (23.8 m/s)^2
= 20.1835 J
Since the initial kinetic energy is 0, the work done on the arrow by the bow string is:
Work = Change in KE = Final KE - Initial KE
= 20.1835 J - 0 J
= 20.1835 J
Therefore, the work done on the arrow by the bow string is 20.1835 Joules.
To determine the force exerted by the archer to pull the bowstring back, we can use the work-force principle. The principle states that work is equal to the force applied multiplied by the distance over which the force is applied.
Work = Force * Distance
Rearranging the equation, we can solve for Force:
Force = Work / Distance
= 20.1835 J / 0.55 m
= 36.6973 N
Therefore, the force exerted by the archer to pull the bowstring back is 36.6973 Newtons.