Answer
a) decrease
b)
$K.E_i=4.70KJ$
$K.E_f=3.82KJ$
Work Step by Step
(a) We know that the kinetic energy of the system should decrease as some of the energy is dissipated in the inelastic collision between the person and the merry-go-round.
(b) We can calculate the initial and final kinetic energies for the system as
$K.E_i=\frac{1}{2}(\frac{1}{2}MR^2)\omega_i^2+\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K.E_i=\frac{1}{4}(155)(2.63)^2(4.03)^2+\frac{1}{2}(59.4)(3.41)^2$
$K.E_i=4.70KJ$
Now $K.E_f=\frac{1}{2}(\frac{1}{2}MR+mR^2)\omega_f^2$
$K.E_f=\frac{1}{2}[\frac{1}{2}(155)(2.63)^2+(59.4)(2.63)^2](2.84)^2$
$K.E_f=3820=3.82KJ$
