91Ó°ÊÓ

i) Mass of ice flake$m=2\text{ g}=2×{10}^{−3}\text{ k²µ}$

ii) hemispherical bowl radius $r=22\text{ c³¾}=22×{10}^{−2}\text{″¾}$

iii) Gravitational acceleration $g=9.8{\text{″¾/s}}^{\text{2}}$

The force of gravity is constant, so the work done can be found by using the formula for the work done in terms of gravitational force and displacement.

Formula:

Gravitational potential energy is given by formula$W_g=mgh$

$W=mgh$, here$h=r$

Change in potential energy$\mathrm{Δ}U=−W_g$

Theforce is vertically downward and has magnitude mg, where mis the mass of the flake, so this reduces to $W=mgh$ where h is the height from which the flake falls. This is equal to the radius r of the bowl.

$W_g=mgr$

Substitute all the value in the above equation.

$\begin{array}{c}{W}_g=2×{10}^{−3}\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×22×{10}^{−2}\text{″¾}\\ W_g=4.31×{10}^{−3}\text{ J}\end{array}$

Hence thework is done on the flake by the gravitational force during the flake’s descent to the bottom of the bowl is $4.31×{10}^{−3}\text{ J}$

The force of gravity is conservative, so the change in gravitational potential energy of the Flake-Earth system is the negative of the work done

$\mathrm{Δ}U=−W_g$

$\begin{array}{c}\mathrm{Δ}U=−mgh\\ =−mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{Δ}U=−2×{10}^{−3}\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×22×{10}^{−2}\text{″¾}\\ \mathrm{Δ}U=−4.31×{10}^{−3}\text{ J}\end{array}$

Hence the change in the potential energy of the flake–Earth system during that descent is $−4.31×{10}^{−3}\text{ J}$

Thepotential energy when the flake is at the top is greater than when it is at the bottom.

Therefore,
$\mathrm{Δ}U=U_{top}−U_{bottom}$

$\begin{array}{l}\mathrm{Δ}U=mgr−0\\ \mathrm{Δ}U=mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{Δ}U=2×{10}^{−3}\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×22×{10}^{−2}\text{″¾}\\ \mathrm{Δ}U=4.31×{10}^{−3}\text{ J}\end{array}$

Hence if that potential energy is taken to be zero at the bottom of the bowl, then value when the flake is released is $4.31×{10}^{−3}\text{ J}$

IfU= 0 at the top,thepotential energy when the flake is at the bottom is less than when it is at the top,

Therefore,

$\begin{array}{l}\mathrm{Δ}U=U_{top}−U_{bottom}\\ \mathrm{Δ}U=0−mgr\\ \mathrm{Δ}U=−mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{Δ}U=−2×{10}^{−3}\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×22×{10}^{−2}\text{″¾}\\ \mathrm{Δ}U=−4.31×{10}^{−3}\text{ J}\end{array}$

Hence if, instead, the potential taken to be zero at the release point, then value when the flake reaches the bottom of the bowl is $−4.31×{10}^{−3}\text{ J}$

The mass is directly proportional to potential energy as well as work so, if mass is doubled then all answer will be doubled.