91Ó°ÊÓ

The total energy of an isolated system remains constant regardless of any internal changes that may occur, whereby energy disappears in one form and reappears in another.

Mass, $m=15.0\text{kg}$

Distance travelled in the earth’s gravitational field, $h_{\text{e}}=5.00\text{m}$

Kinetic energy gained by the drum, $K_{\text{e}}=250.0\text{J}$

Acceleration due to gravity on mars,$g_{\text{m}}=3.71\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.$

In the Earth:

According to law of conservation of energy you can write the following formula.

$$\begin{gathered} K_{\text{e}}+\frac{1}{2}m{v}^2=mgh \\ {K}_{\text{e}}+\frac{1}{2}m{r}^2{\mathrm{Ó\neg }}^2=mgh \\ \frac{1}{2}m{r}^2{\mathrm{Ó\neg }}^2=mgh-K_{\text{e}} \end{gathered}$$

Substitute known values in the above equation.

$\begin{array}{rcl}\frac{1}{2}m{r}^2{\mathrm{Ó\neg }}^2& =& (15.0\text{kg})(9.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.)(5.00\text{m})-250.0\text{J}\\ & =& 485\text{J}\end{array}$

In the mars, the mass and radius of the drum remain unchanged. So the moment of inertia also remains same which implies the kinetic energy of the drum is unaltered. Therefore, kinetic energy of the falling mass also unchanged.

So decrease in potential energy of the mass is Mars is,

$$\begin{gathered} m{g}_{\text{M}}{h}_{\text{M}}=\frac{1}{2}m{\mathrm{Ó\neg }}^2{r}^2+K_{\text{e}} \\ {h}_{\text{M}}=\frac{\frac{1}{2}m{\mathrm{Ó\neg }}^2{r}^2+K_{\text{e}}}{m{g}_{\text{M}}} \end{gathered}$$

Putting known values in the above equation, and you get

$\begin{array}{rcl}{h}_{\text{M}}& =& \frac{485\text{J}+250\text{J}}{\left(15\text{kg}\right)\left(3.71{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.}\right)}\\ & =& 13.2\text{m}\end{array}$

Hence, the required distance is$13.2\text{m}$ would the $15\text{kg}$ mass have to fall to give the same amount of kinetic energy to the drum.

As kinetic energy is same as in earth. Thus, the kinetic energy of the falling mass is,

$$\begin{gathered} \frac{1}{2}m{v}^2=485.0\text{J} \\ {v}^2=\frac{2×485.0\text{J}}{m} \\ v=\sqrt{\frac{2×485.0\text{J}}{15.0\text{kg}}} \end{gathered}$$

role="math" localid="1662226511558" $v=8.04\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Hence, the velocity is$8.04\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ that would the$15\text{kg}$ mass be moving on Mars just as the drum gained$250.0\text{J}$ of kinetic energy.