An archer applies an average force of 200 N to draw the bow string back 1.3 m.

a)How much work did the archer do on the string?

b)How much mechanical potential energy is stored in the stretched bow?
c)How much kinetic energy does the arrow have before it is released?

d)How much potential energy will the arrow have after it leaves the bowstring?

e)If a 0.3 kg arrow is shot from this bow, then how fast will it be moving just after it leaves the bowstring?

I will be happy to critique your work.

OK SO PLEASE EXPLAIN TO ME HOW TO DO THIS

a. work=avgforce*distance

b.PE=above work. KEwhenreleased=InitialPE
d. zero, no PE, it has KE.
e. Initial energy=1/2 m v^2
solve for v.

Idk wat it is

SOMEONE TELL ME HOW TO DO THESE RN PLSPPLSPLSPLSPL

To find the answers to these questions, we need to use the formulas and concepts related to work, potential energy, kinetic energy, and conservation of energy.

a) The work done by the archer on the string can be calculated using the formula:

Work = Force * Distance * Cos(θ)

In this case, the force applied is 200 N, and the distance is 1.3 m. Since the angle is not specified, we can assume it to be 0 degrees, which means Cos(0) = 1. Therefore, the equation simplifies to:

Work = Force * Distance

Substituting the given values, we have:

Work = 200 N * 1.3 m = 260 J

So, the archer did 260 Joules of work on the string.

b) The mechanical potential energy stored in the stretched bow can be calculated using the formula:

Potential Energy = (1/2) * k * x^2

Here, k represents the spring constant of the bow (which is unknown in this case), and x represents the distance the bowstring is stretched. Since we are not given the spring constant, we cannot calculate the exact potential energy. However, we know that the potential energy is directly proportional to the distance the bowstring is stretched squared.

c) The kinetic energy of the arrow just before it is released can be calculated using the formula:

Kinetic Energy = (1/2) * mass * velocity^2

Since only the mass of the arrow is given (0.3 kg), we need to find the velocity. To do this, we can use the concept of conservation of energy. The work done by the archer in drawing the bowstring is converted into the potential energy stored in the bow. The release of the bow transfers this potential energy into the kinetic energy of the arrow.

Using the equation for work from part (a), we know that this work is equal to the potential energy stored in the bow. We can set them equal to each other:

Potential Energy = Work = 260 J

Now, since potential energy is converted into kinetic energy, we can write the equation:

Potential Energy = Kinetic Energy

Substituting the given value of potential energy, we have:

260 J = (1/2) * 0.3 kg * velocity^2

Rearranging the equation and solving for velocity, we find:

velocity^2 = (260 J * 2) / (0.3 kg)
velocity^2 = 1733.33 m^2/s^2

Taking the square root of both sides:

velocity ≈ 41.63 m/s

So, the arrow has a speed of approximately 41.63 m/s just after it leaves the bowstring.

d) The potential energy of the arrow after it leaves the bowstring can be calculated using the formula:

Potential Energy = (1/2) * k * x^2

However, we still do not know the spring constant of the bow (k) or the distance the bowstring is stretched (x). So, without more information, we cannot determine the potential energy of the arrow after it leaves the bowstring.

e) The speed of the arrow just after it leaves the bowstring is calculated as 41.63 m/s, as determined in part (c).