Answer:
Explanation:
Given
Radius of Pulley r=12 cm
mass of block m=60 gm
mass of Pulley M=430 gm
Block descend h=50 cm
Applying Conservation of Energy
Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block
i.e.
[tex]mgh=\frac{1}{2}I\omega ^2+\frac{1}{2}mv^2[/tex]
where I=moment of inertia
[tex]I=mr^2[/tex]
and for rolling [tex]\omega =\frac{v}{r}[/tex]
[tex]mgh=\frac{1}{2}Mv^2+\frac{1}{2}mv^2[/tex]
[tex]v^2=\frac{2mgh}{m+M}[/tex]
[tex]v=\sqrt{\frac{2mgh}{m+M}}[/tex]
[tex]v=\sqrt{\frac{2\times 60\times 9.8\times 0.5}{430+60}}[/tex]
[tex]v=\sqrt{\frac{60\times 9.8}{490}}[/tex]
[tex]v=\sqrt{1.2}[/tex]
[tex]v=1.095 m/s[/tex]
