91Ó°ÊÓ

i) Mass of book$m=2\text{ k²µ}$

ii) Distance stands on the ground$\left(D\right)=10\text{″¾}$

iii) Stretched hand at distance $\left(d\right)=1.50\text{″¾}$

iv) Gravitational acceleration$\left(D\right)=9.8{\text{″¾/s}}^{\text{2}}$

By using the concept of potential energy, we can find gravitational work. Gravitational work is nothing but the potential energy due the gravitational force.

i.e.$\mathit{U}\mathbf{=}{\mathit{W}}_{\mathbf{g}}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$

Formula:

Gravitational potential energy is given by theformula

$\begin{array}{c}\mathbf{U}\mathbf{=}{\mathbf{W}}_{\mathbf{g}}\\ \mathbf{=}\mathbf{m}\mathbf{g}\mathbf{h}\end{array}$

We can find the height as below

$h=D−d$

Substitute all the value in the above equation.

$\begin{array}{c}h=10\text{″¾}−1.50\text{″¾}\\ h=8.5\text{″¾}\end{array}$

$h=8.5\text{″¾}$(Downward direction same as$F_g$)

Work depends ontheinitial and final position.

$W_g=mgh$

Substitute all the value in the above equation.

$\begin{array}{c}{W}_g=2\text{ k²µ}×9.80\text{″¾}/\text{s}×8.50\text{″¾}\\ =166.6\text{ J}\\ W_g=167\text{ J}\end{array}$

Work $W_g$does the gravitational force do on the book as it drops from her hands is$167\text{ J}$

We must calculate change in potential energy, so that

$\mathrm{Δ}U=mgh$But here h is$h=d−D$i.e. final height minus initial height

$\mathrm{Δ}U=mg(d−D)$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{Δ}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×(−8.50\text{″¾})\\ \mathrm{Δ}U=−167\text{ J}\end{array}$

Change$\mathrm{Δ}U$ in the gravitational potential energy of the book–Earth system during the drop If the gravitational potential energy of$U$ that system is taken to be zero at ground level,$−167\text{ J}$

Initial potential energy,

$U=mgD$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×\left(10\text{″¾}\right)\\ U=196\text{ J}\end{array}$

Potential energy$U$whenthebook is released is$196\text{ J}$

Final potential energy,

$U=mgd$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×\left(1.50\text{″¾}\right)\\ U=29\text{ J}\end{array}$

Potential energy$U$whenthebook reaches her hand is$29\text{ J}$

Work does not depend on initial value of potential energy. So that

$W_g=mgh$

Substitute all the value in the above equation.

$\begin{array}{c}{W}_g=2\text{ k²µ}×9.80\text{″¾}/\text{s}×8.50\text{″¾}\\ =166.6\text{ J}\\ W_g=167\text{ J}\end{array}$

Work $W_g$due to gravitational force is$167\text{ J}$

Change in potential energy

$\mathrm{Δ}U=−W_g=−mg(D−d)$

$W_g=mg(d−D)$

Substitute all the value in the above equation

$\begin{array}{c}\mathrm{Δ}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×(−8.50\text{″¾})\\ \mathrm{Δ}U=−167\text{ J}\end{array}$

Change in potential energy is$−167\text{ J}$

Initial potential energy,

$U=mgD+U_0$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×\left(10\text{″¾}\right)+100\text{ J}\\ U=296\text{ J}\end{array}$

Potential energy$U$at the release point is 296 J.

Final potential energy,

$U=mgd+U_0$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{ k²µ}×9.80\text{″¾}/{\text{s}}^{\text{2}}×\left(1.5\text{″¾}\right)+100\text{ J}\\ U=129\text{ J}\end{array}$

Potential energy$U$at her hands is$129\text{ J}$