91Ó°ÊÓ

1. Wheel radius is $0.20\text{m}$
2. Rotational inertia is$0.40\text{kg}-{\text{m}}^2$
3. Mass of box is$6.0\text{kg}$
4. Kinetic energy of box is $6.0\text{J}$

Use the formula in terms of inertia and angular velocity to find the rotational kinetic energy. Then, using the law of conservation of energy, find the height through which the box has fallen.

Formulae are as follow:

1. $K=\frac{1}{2}m{v}^2$
2. $K_{rot}=\frac{1}{2}I{\mathrm{Ó\neg }}^2$
3. $v=r\mathrm{Ó\neg }$

Where,

K is kinetic energy, m is mass, r is radius, I is moment of inertia, v is velocity and is angular velocity.

First, find the velocity from kinetic energy as follows:

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ 6=\frac{1}{2}6v^2 \\ v=1.41\text{m}/\text{s} \end{gathered}$$

Now, find angular velocity as follows:

$$\begin{gathered} v=r\mathrm{Ó\neg } \\ 1.41=0.2\mathrm{Ó\neg } \\ \mathrm{Ó\neg }=7.05\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right. \end{gathered}$$

Now, rotational kinetic energy is as follows:

$$\begin{gathered} K{E}_{rotational}=0.5×I{\mathrm{Ó\neg }}^2 \\ K{E}_{rotational}=0.5×0.40×7.{05}^2 \\ K{E}_{rotational}=9.9405\text{J} \end{gathered}$$

In one significant figure,

$K{E}_{rotational}=10\text{J}$

Hence, wheel’s rotational kinetic energy is $10\text{J}$

Now, conservation of energy,

$$\begin{gathered} K_i+U_i=K_f+U_f \\ {K}_i+U_i=\left(K_{translation}+K_{rotational}\right)+U_f \\ 0+0=\left(6+10\right)-6×9.8×h \\ h=0.27\text{m} \end{gathered}$$

Hence, distance through which the box has fallen is$0.27\text{m}$

Therefore, using the formula for rotational kinetic energy and law of conservation of energy, the distance through which the box has fallen can be found.