Answer
(a) $1.19J$
(b) $1.19J$
(c) $2.38J$
Work Step by Step
(a) We know that translational kinetic energy is given as
$K.E_t=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K.E_t=\frac{1}{2}(1.2Kg)(1.41m/s)^2$
$K.E_t=1.19J$
(b) Rotational kinetic energy is given as
$K.E_{rot}=\frac{1}{2}I\omega^2$
We plug in the known values to obtain:
$K.E_{rot}=\frac{1}{2}(mr^2)\omega^2$
$K.E_{rot}=\frac{1}{2}m(\omega r)^2$
$K.E_{rot}=\frac{1}{2}(1.2Kg)(1.41m/s)^2$
$K.E_{rot}=1.19J$
(c) The total kinetic energy is given as
$K.E_{total}=K.E_t+K.E_{rot}$
We plug in the known values to obtain:
$K.E_{total}=2.38J$
