Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.
The speed of arrow leaving the bow is 10.01 m/s.
Given data:
The distance drawn by bow is, d = 0.29 m.
The mass of arrow is, m = 0.094 kg.
The value of spring constant is, k = 112 N/m.
The spring potential energy stored during the loading of bow is converted into kinetic energy of arrow. Then,
Spring potential energy = Kinetic energy
[tex]U=KE\\\dfrac{1}{2}kd^2 =\dfrac{1}{2}mv^2\\kd^2=mv^2\\v=\sqrt{ \dfrac{kd^2}{m}}\\v=\sqrt{ \dfrac{112 \times 0.29^2}{0.094}}\\v = 10.01 \;\rm m/s[/tex]
Thus, we can conclude that the speed of arrow leaving the bow is 10.01 m/s.
Learn more about spring potential energy here:
https://brainly.com/question/18167148?referrer=searchResults
