You drop a$\mathbf{2}\mathbf{.00}\mathbf{\text{\hspace{0.17em}Kg}}$book to a friend who stands on the ground at distance$\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$below. If your friend’s outstretched hands are at distance$\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.50}\mathbf{\text{\hspace{0.17em}m}}$above (fig.8-30), (a) how much work${\mathit{W}}_{\mathbf{g}}$does the gravitational force do on the book as it drops to her hands? (b) What is the change$\mathit{\Delta }\mathit{U}$in the gravitational potential energy of the book- Earth system during the drop? If the gravitational potential energy U of the system is taken to the zero at ground level, what is U (c) when the book is released and (d) when it reached her hands? Now take U to be 100 J at ground level and again find (e) ${\mathit{W}}_{\mathbf{g}}$,(f) $\mathit{\Delta }\mathit{U}$(g) U at the release point, and (h) U at her hands.

i) Mass of book$m=2\text{\hspace{0.17em}kg}$

ii) Distance stands on the ground$\left(D\right)=10\text{\hspace{0.17em}m}$

iii) Stretched hand at distance $\left(d\right)=1.50\text{\hspace{0.17em}m}$

iv) Gravitational acceleration$\left(D\right)=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$

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.

$\begin{array}{c}h=10\text{\hspace{0.17em}m}-1.50\text{\hspace{0.17em}m}\\ h=8.5\text{\hspace{0.17em}m}\end{array}$

$h=8.5\text{\hspace{0.17em}m}$(Downward direction same as$F_g$)

Work depends ontheinitial and final position.

$W_g=mgh$

Substitute all the value in the above equation.

$\begin{array}{c}{W}_g=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/\text{s}\times 8.50\text{\hspace{0.17em}m}\\ =166.6\text{\hspace{0.17em}J}\\ W_g=167\text{\hspace{0.17em}J}\end{array}$

Work $W_g$does the gravitational force do on the book as it drops from her hands is$167\text{\hspace{0.17em}J}$

We must calculate change in potential energy, so that

$\Delta U=mgh$But here h is$h=d-D$i.e. final height minus initial height

$\Delta U=mg(d-D)$

Substitute all the value in the above equation.

$\begin{array}{c}\Delta U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times (-8.50\text{\hspace{0.17em}m})\\ \Delta U=-167\text{\hspace{0.17em}J}\end{array}$

Change$\Delta U$ 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,$-167\text{\hspace{0.17em}J}$

Initial potential energy,

$U=mgD$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(10\text{\hspace{0.17em}m}\right)\\ U=196\text{\hspace{0.17em}J}\end{array}$

Potential energy$U$whenthebook is released is$196\text{\hspace{0.17em}J}$

Final potential energy,

$U=mgd$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(1.50\text{\hspace{0.17em}m}\right)\\ U=29\text{\hspace{0.17em}J}\end{array}$

Potential energy$U$whenthebook reaches her hand is$29\text{\hspace{0.17em}J}$

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

$W_g=mgh$

Substitute all the value in the above equation.

$\begin{array}{c}{W}_g=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/\text{s}\times 8.50\text{\hspace{0.17em}m}\\ =166.6\text{\hspace{0.17em}J}\\ W_g=167\text{\hspace{0.17em}J}\end{array}$

Work $W_g$due to gravitational force is$167\text{\hspace{0.17em}J}$

Change in potential energy

$\Delta U=-W_g=-mg(D-d)$

$W_g=mg(d-D)$

Substitute all the value in the above equation

$\begin{array}{c}\Delta U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times (-8.50\text{\hspace{0.17em}m})\\ \Delta U=-167\text{\hspace{0.17em}J}\end{array}$

Change in potential energy is$-167\text{\hspace{0.17em}J}$

Initial potential energy,

$U=mgD+U_0$

Substitute all the value in the above equation.

$\begin{array}{c}U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(10\text{\hspace{0.17em}m}\right)+100\text{\hspace{0.17em}J}\\ U=296\text{\hspace{0.17em}J}\end{array}$

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.

$\begin{array}{c}U=2\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(1.5\text{\hspace{0.17em}m}\right)+100\text{\hspace{0.17em}J}\\ U=129\text{\hspace{0.17em}J}\end{array}$

Potential energy$U$at her hands is$129\text{\hspace{0.17em}J}$