Answer
(a) $0J, 113J,113J$
(b) $113J, 226J, 113J$
Work Step by Step
(a) We can find the required kinetic energies as follows:
$K.E_i=(\frac{1}{2})mv_i^2$
$\implies K.E_i=\frac{1}{2}m(0)^2=0J$
Now the final kinetic energy is given as
$K.E_f=-mgy$
We plug in the known values to obtain:
$K.E_f=-(5.76)(9.81)(-2.0)=113J$
Now the change in kinetic energy can be determined as
$\Delta K.E=K.E_f-K.E_i$
$\Delta K.E=113-0=113J$
(b) As $K.E_i=113J$
Now the final kinetic energy can be determined as
$K.E_f=K.E_i+mg(y_i-y_f)$
We plug in the known values to obtain:
$K.E_f=113+(5.76)(9.81)[-2.00-(-4.00)]$
$K.E_f=226J$
$\Delta K.E=K.E_f-K.E_i$
We plug in the known values to obtain:
$\Delta K.E=226-113=113J$
