Answer:
[tex]v = 2.37 m/s[/tex]
Explanation:
As we know that here larger bucket will move down and smaller bucket will move up by same distance
So by energy conservation we can say that decrease potential energy = gain in kinetic energy of the system
So we will have
[tex]m_1gh - m_2gh = \frac{1}{2}(m_1 + m_2) v^2 + \frac{1}{2}I\omega^2[/tex]
also we know that string moves over the pulley without slipping
so we will have
[tex]\omega = \frac{v}{R}[/tex]
now we have
[tex](19.9 - 13.7)(9.81)(1.75) = \frac{1}{2}(19.9 + 13.7)v^2 + \frac{1}{2}(\frac{1}{2}m_pR^2)(\frac{v}{R})^2[/tex]
[tex]106.44 = 16.8 v^2 + \frac{1}{4}(8.73)v^2[/tex]
[tex]106.44 = 18.98 v^2[/tex]
[tex]v = 2.37 m/s[/tex]
This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.
The speed with which the larger bucket hits the ground will be "2.24 m/s".
Applying the law of conservation of energy to this situation we have:
Change in Potential Energy = Change in Kinetic Energy
[tex]m_1gh-m_2gh=\frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2+\frac{1}{2}I\omega^2[/tex]
where,
m₁ = mass of larger bucket = 19.9 kg
m₂ = mass of smaller bucket = 13.7 kg
h = height = 1.75 m
v = speed of buckets = ?
r = radius of pulley = 0.15 m
mp = mass of pulley = 8.73 kg
I = moment of inertia of pulley = [tex]\frac{1}{2}m_p r^2=\frac{1}{2}(8.73\ kg)(0.15\ m)^2=0.098\ kg.m^2[/tex]
ω = angular speed = [tex]\frac{v}{r}=\frac{v}{0.15\ m}[/tex]
Therefore,
[tex](19.9\ kg-13.7\ kg)(9.81\ m/s^2)(1.75\ m)=\frac{1}{2}(19.9\ kg+13.7\ kg)v^2+\frac{1}{2}(0.098)(\frac{v^2}{(0.15\ m)^2})\\\\106.44=16.8\ v^2+4.365\ v^2\\\\v=\sqrt{\frac{106.44}{21.165}}[/tex]v = 2.24 m/s
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
