Answer:
a. The arrow leaves the bow at 30 m/s.
b. The maximum height of the arrow is 46 m.
Step-by-step explanation:
Hi there!
a. As the string is being stretched the arrow acquires elastic potential energy and when it is released, all the elastic potential energy of the arrow will be transformed into kinetic energy (since no energy is dissipated by friction). Then, the initial elastic potential energy (EPE) of the arrow is equal to its final kinetic energy (KE):
EPE = KE
1/2 · k · x² = 1/2 · m · v²
Where:
k = spring constant.
x = stretching distance of the string.
m = mass of the arrow.
v = speed.
The spring constant can be calculated using Hooke´s law (the force needed to stretch or compress a spring is proportional to the stretching or compressing distance (i.e: the more you compress or stretch the spring, the more force you need to apply)):
Mathematically, it is expressed as follows:
F = kx
F/x = k
Then, replacing k in the equation:
1/2 · k · x² = 1/2 · m · v²
1/2 · F/x · x² = 1/2 · m · v²
1/2 · F · x = 1/2 · m · v²
Solving for v:
F · x / m = v²
v = √(F · x / m)
v = √ (201 N · 1.3 m / 0.30 kg)
v = 30 m/s
The arrow leaves the bow at 30 m/s
b. If the arrow is shot straight up, the initial kinetic energy (or elastic potential energy) will be converted into gravitational potential energy. At the maximum height, the kinetic energy of the arrow is zero (v = 0, KE = 0) meaning that all the initial kinetic energy of the arrow was converted into gravitational potential energy (PE). Then, the final potential energy has to be equal to the initial kinetic energy because the total energy is conserved (i.e. it remains constant).
PE = KE
m · g · h = 1/2 · m · v²
Where:
g = acceleration due to gravity (9.8 m/s²)
h = height.
Solving for h:
h = 1/2 · v²/g
h = 1/2 · (30 m/s)² / 9.8 m/s²
h = 46 m (44 m without rounding the velocity).
The maximum height of the arrow is 46 m.