An archer puts a 0.28 kg arrow to the bow string. An average force of 200 N is exerted to draw the sting back 1.1 m. a) Assuming no frictional loss, what speed does the arrow leave the bow? b) If the arrow is shot straight up, how high does it rise?
My thoughts: KE = 1/2mv^2. You are given mass but how do you find velocity? Once velocity is given KE = PE = mgh.
(a) Work done drawing the arrow back
= 200 N x 1.1 m
=(1/2) M V^2
Solve for V
(b) Knowing V, the maximum height attained (h) is given by
(V^2/2g) = h
To find the velocity of the arrow leaving the bow, we can use the principle of conservation of energy. The potential energy stored in the bowstring is converted into kinetic energy as the arrow is released.
a) First, let's calculate the potential energy stored in the bowstring. The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the arrow, g is the acceleration due to gravity, and h is the vertical distance through which the bowstring is drawn.
In this case, the mass of the arrow is 0.28 kg and the height through which the bowstring is drawn is 1.1 m. The acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the potential energy is given by PE = (0.28 kg)(9.8 m/s^2)(1.1 m).
Next, we can equate the potential energy stored in the bowstring to the kinetic energy of the arrow just as it leaves the bow. The kinetic energy (KE) is given by the equation KE = 1/2mv^2, where v is the velocity of the arrow.
Since energy is conserved, we can write PE = KE. Therefore, (0.28 kg)(9.8 m/s^2)(1.1 m) = 1/2(0.28 kg)v^2.
Now, we can solve for v. Rearranging the equation, we get v^2 = (2 × (0.28 kg)(9.8 m/s^2)(1.1 m)) / 0.28 kg.
By simplifying the equation, we find v^2 = 2(9.8 m/s^2)(1.1 m). Now, we can calculate v by taking the square root of both sides: v = √(2(9.8 m/s^2)(1.1 m)).
b) To find the maximum height reached by the arrow when it is shot straight up, we can use the same principle of conservation of energy. This time, the potential energy at maximum height is equal to the kinetic energy just as the arrow is released.
Let's denote the maximum height reached by the arrow as H. Using the equation PE = KE, we have mgh = 1/2mv^2, where m is the mass of the arrow, g is the acceleration due to gravity, h is the maximum height, and v is the final velocity of the arrow.
Canceling out the mass 'm', we get gh = 1/2v^2. Now, we can rearrange the equation to solve for the maximum height H: H = (1/2v^2) / g.
Plugging in the value of v, which we calculated in part a, and the acceleration due to gravity (g ≈ 9.8 m/s^2), we can find the maximum height H.