An arrow with a mass of 0.12 kg is pulled back in a bow. After being held still, it is let go, moving at 23m/s after the 0.51meters in which the arrow is in contact with the bow string? What work was done on the arrow by th bow string? What force was exerted by the archer to pull the bowstring back?
Hello, why is there a 1/2 in your kinetic formula if you are using Fmax and delta x? W= (Fmax)(delta x) and Ek =1/2 m(v^2)
To find the work done on the arrow by the bowstring, we can use the work-energy principle:
Work done = Final kinetic energy - Initial kinetic energy
The initial kinetic energy is zero, as the arrow is held still before being released. The final kinetic energy can be calculated using the formula:
Kinetic energy = 1/2 * mass * velocity^2
Given:
Mass of the arrow (m) = 0.12 kg
Final velocity of the arrow (v) = 23 m/s
Final kinetic energy = 1/2 * 0.12 kg * (23 m/s)^2
Now we can calculate the work done:
Work done = Final kinetic energy - Initial kinetic energy = 1/2 * 0.12 kg * (23 m/s)^2 - 0
To find the force exerted by the archer to pull the bowstring back, we can use Newton's second law of motion:
Force = Mass * Acceleration
We can calculate the acceleration of the arrow using the formula:
Acceleration = Change in velocity / Time
Given:
Change in velocity (Δv) = 23 m/s
Time (t) = 0.51 m
Acceleration = Δv / t = 23 m/s / 0.51 m
Now we can calculate the force exerted:
Force = Mass * Acceleration = 0.12 kg * (23 m/s / 0.51 m)
To find the work done on the arrow by the bowstring, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
First, let's find the initial kinetic energy of the arrow. The initial velocity is 0 because the arrow is held still in the bow. Therefore, the initial kinetic energy is 0.
Next, let's find the final kinetic energy of the arrow. The arrow is moving at a final velocity of 23 m/s. The mass of the arrow is 0.12 kg. The formula for kinetic energy is K.E. = (1/2)mv^2. Plugging in the values, we get K.E. = (1/2)(0.12 kg)(23 m/s)^2 = 29.988 J (rounded to three decimal places).
Since the initial kinetic energy is 0, the work done on the arrow by the bowstring is equal to the change in kinetic energy. Therefore, the work done on the arrow by the bow string is 29.988 Joules (rounded to three decimal places).
Now, let's determine the force exerted by the archer to pull the bowstring back. The work done by a force can be calculated using the formula W = Fd, where W is the work done, F is the force, and d is the distance over which the force is applied.
In this case, the work done by the archer is equal to the work done on the arrow, which we found to be 29.988 Joules. The distance, d, over which this force is applied is given as 0.51 meters.
Substituting the values into the formula, we have 29.988 J = F * 0.51 m. Solving for F, we get F = 58.8 N (rounded to one decimal place).
Therefore, the force exerted by the archer to pull the bowstring back is approximately 58.8 Newtons.
In this problem, it is best to assume that the force applied to the arrow is proportional to the distance that the arrow is pulled back. Thus an uneven force is applied, starting at Fmax and ending at zero when the string is straight up and down.
The kinetic energy given to the arrow equals the work done by the bowstring in pushing the arrow forward. This equals (1/2) Fmax * X, where X = 0.51 m
Fmax = (1/2) M V^2/(X/2) = M V^2/X
Solve fr Fmax