Answer
The rolling wheel: $KE=54J$ and $h_{max}=2.76m$
The sliding wheel: $KE=36J$ and $h_{max}=1.84m$
Work Step by Step
a) The rolling wheel's total kinetic energy: $$KE=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$$
The wheel has a rolling motion, so $\omega=\frac{v}{R}$. Its moment of inertia has the formula $I=\frac{1}{2}mR^2$
Therefore, its rotational KE is $\frac{1}{2}I\omega^2=\frac{1}{4}mR^2\frac{v^2}{R^2}=\frac{1}{4}mv^2$
$$KE=\frac{3}{4}mv^2=\frac{3}{4}(2kg)(6m/s)^2=54J$$
a) The sliding wheel's total kinetic energy: $$KE=\frac{1}{2}mv^2=\frac{1}{2}(2kg)(6m/s)^2=36J$$
b) According to the principle of conservation of mechanical energy, $$E_f=E_0$$ $$KE_f+PE_f=KE_0+PE_0$$
For both wheels, they start off from the ground, so their initial potential energy $PE_0=0$. As they reach their maximum height, their speed is zero, $KE_f=0$. Therefore,
$$PE_f=KE_0$$ $$mgh_f=KE_0$$ $$h_{max}=\frac{KE_0}{mg}=\frac{KE_0}{19.6}$$
a) The rolling wheel has $KE_0=54J$, so $h_{max}=2.76m$
b) The sliding wheel has $KE_0=36J$, so $h_{max}=1.84m$
