Answer
(a) $18.4J$
(b) $5.26J$
(c) $13.13J$
Work Step by Step
(a) We can find the total kinetic energy as follows:
$U_i+K_i=U_f+K_f$
We know that at the top, the initial kinetic energy is zero and at the ground the gravitational potential energy is zero. Thus, the above equation becomes
$U_i=K_f$
$\implies K_f=mgh$
We plug in the known values to obtain:
$K_f=(2.5Kg)(9.81m/s^2)(0.75m)$
$K_f=18.4J$
(b) We can find the rotational kinetic energy as follows:
$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$mgh=\frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2})$
This simplifies to:
$v^2=\frac{10}{7}gh$.........eq(1)
Now $K_r=\frac{1}{2}I\omega^2$
$\implies K_r=\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2})$
$\implies K_r=\frac{1}{5}mv^2$
from eq(1) $v^2=\frac{10}{7}gh$
$\implies K_r=\frac{1}{5}m(\frac{10}{7}gh)$
$K_r=\frac{2}{7}mgh$
We plug in the known values to obtain:
$K_r=\frac{2}{7}(2.5Kg)(9.81m/s^2)(0.75m)$
$K_r=5.26J$
(c) We can find the required translational kinetic energy as
$K_t=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K_t=\frac{1}{2}m(\frac{10}{7}gh)$
We plug in the known values to obtain:
$K_t=\frac{5}{7}mgh$
We plug in the known values to obtain:
$K_t=\frac{5}{7}(9.8m/s^2)(0.75m)$
$K_t=13.13J$
