Question

In Fig. 8-32, a ice flake is released from the edge of a hemisphere bowl whose radius ris $\mathbf{22}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$. The flake-bowl contact is frictionless. (a) How much work is done on the flake by the gravitational force during the flake’s descent to the bottom of the bowl? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl? (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (b) increase, decrease, or remain the same?

Short answer

a) Work is done on the flake by the gravitational force during the flake’s descent to the bottom of the bowl is $4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

b) Change in the potential energy of the flake–Earth system during that descent is $-4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

c) If that potential energy is taken to be zero at the bottom of the bowl, then value when the flake is released is $4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

d) If, instead, the potential taken to be zero at the release point, then value when the flake reaches the bottom of the bowl is $-4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

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

Step-by-step solution

Given

i) Mass of ice flake$m=2\text{\hspace{0.17em}g}=2\times {10}^{-3}\text{\hspace{0.17em}kg}$

ii) hemispherical bowl radius $r=22\text{\hspace{0.17em}cm}=22\times {10}^{-2}\text{\hspace{0.17em}m}$

iii) Gravitational acceleration $g=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$

To understand the concept

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$W_g=mgh$

(a) Calculate how much work is done on the flake by the gravitational force during the flake’s descent to the bottom of the bowl

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

Change in potential energy$\Delta U=-W_g$

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.

$\begin{array}{c}{W}_g=2\times {10}^{-3}\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 22\times {10}^{-2}\text{\hspace{0.17em}m}\\ W_g=4.31\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

Hence thework is done on the flake by the gravitational force during the flake’s descent to the bottom of the bowl is $4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

(b) Calculate the change in the potential energy of the flake-Earth system during that descent 

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

$\Delta U=-W_g$

$\begin{array}{c}\Delta U=-mgh\\ =-mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\Delta U=-2\times {10}^{-3}\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 22\times {10}^{-2}\text{\hspace{0.17em}m}\\ \Delta U=-4.31\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

Hence the change in the potential energy of the flake–Earth system during that descent is $-4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

(c) Calculate the value of potential energy when the flake is released if potential energy is taken to be zero at the bottom of the bowl

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

Therefore,
$\Delta U=U_{top}-U_{bottom}$

$\begin{array}{l}\Delta U=mgr-0\\ \Delta U=mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\Delta U=2\times {10}^{-3}\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 22\times {10}^{-2}\text{\hspace{0.17em}m}\\ \Delta U=4.31\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

Hence if that potential energy is taken to be zero at the bottom of the bowl, then value when the flake is released is $4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

(d) Calculate the potential energy when the flake reaches the bottom of the bowlif the potential energy is taken to be zero at the release point

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

Therefore,

$\begin{array}{l}\Delta U=U_{top}-U_{bottom}\\ \Delta U=0-mgr\\ \Delta U=-mgr\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\Delta U=-2\times {10}^{-3}\text{\hspace{0.17em}kg}\times 9.80\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 22\times {10}^{-2}\text{\hspace{0.17em}m}\\ \Delta U=-4.31\times {10}^{-3}\text{\hspace{0.17em}J}\end{array}$

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 $-4.31\times {10}^{-3}\text{\hspace{0.17em}J}$

(e) Figure out if the magnitudes of the answers to (a) through (b) increase, decrease, or remain the same if the mass of the flake were doubled, would

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