Question

A mass of 2000 kg is currently being lowered at a velocity 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is $150kg-m^2$. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is

Answer 1

$E_{initial}-E_{brake}=E_{final}$
$\frac{1}{2}I{\omega }_1^2+\frac{1}{2}m{v}_1^2+mg{H}_1-E_{brake}$
$=\frac{1}{2}I{\omega }_2^2+\frac{1}{2}m{v}_2^2+mg{H}_2$
$E_{brake}=\frac{1}{2}I{\omega }_1^2+\frac{1}{m{v}_1^2}+mg(H_1-H_2)$
$=h=H_1-H_2=0.5m(\frac{1}{2}I{\omega }_2^2=0;\frac{1}{2}m{V}_2^2=0)$
$E_{brake}=\frac{1}{2}I{\omega }_1^2+\frac{1}{2}m{v}_1^2+mg(H_1-H_2)$
$h=H_1-H_2=0.5m(\frac{1}{2}I{w}_2^2=0;\frac{1}{2}m{V}_2^2=0)$
$=\frac{\frac{1}{2}\times 150\times 2^2+2000\times 9.81\times 0.5+\frac{1}{2}\times 2000\times 2^2}{1000}$
$=14.11kJ$