Respuesta :
Answer:
(D) 22 J
Explanation:
Given:
radius of the pulley, [tex]r=3\,cm[/tex]
rotational inertia of the pulley, [tex]I=4.5\times 10^{-3}\,kg.m^{-2}[/tex]
mass on one side of the rope, [tex]m_1=2\,kg[/tex]
mass on the other side of the rope, [tex]m_2=4\,kg[/tex]
velocity of the mass [tex]m_2[/tex], [tex]v=2\,m.s^{-1}[/tex]
Angular velocity of the pulley:
[tex]\omega= \frac{v}{r}[/tex]
[tex]\omega=\frac{2}{3\times 10^{-2}}\,rad.s^{-1}[/tex]
We know for a rotating body,
[tex]KE=\frac{1}{2} I.\omega^2[/tex]
putting the respective values
[tex]KE_p=\frac{1}{2} \times 4.5\times 10^{-3}\times (\frac{200}{3})^2[/tex]
[tex]KE_p= 10\,J[/tex]
Now, the kinetic energy of the blocks :
[tex]KE_b=\frac{1}{2} (m_1+m_2).v^2[/tex]
[tex]KE_b=\frac{1}{2}\times (2+4)\times 2^2[/tex]
[tex]KE_b=12\,J[/tex]
∴Total kinetic energy
[tex]KE=KE_p+KE_b[/tex]
[tex]KE=10+12[/tex]
[tex]KE=22\,J[/tex]
