A 25 gram arrow fired from a bow has a speed of 80m/s. The energy in the arrow came from work done by the bow that acted over a 1m "draw" distance. What average force acted on the arrow over this distance
mass=0.025kg
speed=80m/s
avg force 2N
Please show equations or calculations used in obtaining your answer so that we can help you check your work.
To find the average force acting on the arrow, we can use the work-energy principle. According to the principle, the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the arrow is given by the formula: KE = 0.5 * mass * (speed)^2
Plugging in the values given: KE1 = 0.5 * 0.025 kg * (80 m/s)^2
The final kinetic energy of the arrow is zero since it comes to a stop after being fired. Therefore, KE2 = 0.
The work done by the bow is equal to the change in kinetic energy: W = KE2 - KE1 = - KE1
The work done by the bow can also be calculated using the formula: W = force * distance
Plugging in the values given: W = force * 1 m
Since W = - KE1, we can set the two equations equal to each other:
force * 1 m = - KE1
Now, rearranging the equation to solve for the force:
force = - KE1 / 1 m
Plugging in the values, we get:
force = - (0.5 * 0.025 kg * (80 m/s)^2) / 1 m
Simplifying the equation:
force = - 0.5 * 0.025 kg * (80 m/s)^2
Calculating the force:
force = - 20 N
The average force is given by the magnitude of the force. Taking the absolute value, we get:
average force = |force| = |-20 N| = 20 N
Therefore, the average force acted on the arrow over the 1 m draw distance is 20 N.