Chapter 8, Problem 82PQ

(a)

To determine

The gravitational potential energy of the graph-Earth system at the grape’s initial position.

Answer to Problem 82PQ

The gravitational potential energy of the graph-Earth system at the grape’s initial position is $6.47\times {10}^{-3}\text{J}$ .

Explanation of Solution

The grapes position at different heights is shown below.

Write the expression for the radius of bowl.

  $r=\frac{d}{2}$                                                                                                                         (I)

Here, $r$ is the radius of bowl, $d$ is the diameter of the bowl.

Write the expression for the gravitational potential energy.

  $U=mgh$                                                                                                                  (II)

Here, $U$ is the gravitational potential energy, $m$ is the mass of grape, $g$ is the acceleration due to gravity and $h$ is the height of grape from bottom of bowl.

Conclusion:

Initially grape is resting at upper edge. Thus, initial height is equal to radius of bowl.

Substitute $44.0\text{cm}$ for $d$ in equation (I) to get $r$.

  $\begin{array}{c}r=\frac{0.44\text{cm}}{2}\\ =0.220\text{cm}\end{array}$

Substitute $0.220\text{cm}$ for $r$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $3.00\text{g}$ for $m$ in equation(II) to get potential energy at initial point.

  $\begin{array}{c}{U}_1=\left(3.00\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(22.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =6.47\times {10}^{-3}\text{J}\end{array}$

Here, $U_1$ is the potential energy when the grape is at top of bowl.

Therefore, the gravitational potential energy of the graph-Earth system at the grape’s initial position is $6.47\times {10}^{-3}\text{J}$ .

(b)

To determine

The kinetic energy of the grape when it reaches the bottom of the bowl.

Answer to Problem 82PQ

The kinetic energy of the grape when it reaches the bottom of the bowl is $6.47\times {10}^{-3}\text{J}$ .

Explanation of Solution

Take the bottom of the bowl as $h=0$ point.

Write conservation of energy equation as the grape moves from top of the bowl to bottom of bowl.

  $U_1+K_1=U_3+K_3$                                                                                                    (III)

Here, $K_1$ is the kinetic energy at top of bowl, $U_3$ is the potential energy at bottom of bowl and $K_3$ is the kinetic energy at bottom of bowl.

Conclusion:

In problem it is given that initially the grape is at rest at upper edge of bowl. At bottom of bowl potential energy is zero, since $h=0$. Potential energy at upper edge of the ball is obtained as $6.47\times {10}^{-3}\text{J}$.

Substitute $0\text{J}$ for $K_1$ and $0\text{J}$ for $U_3$ in equation (III) to get $K_3$.

  $\begin{array}{l}{U}_1+0\text{J}=0\text{J}+K_3\\ K_3=U_1\end{array}$

Substitute $6.47\times {10}^{-3}\text{J}$ for $U_1$ in above equation to get $K_3$.

  $K_3=6.47\times {10}^{-3}\text{J}$

Therefore, the kinetic energy of the grape when it reaches the bottom of the bowl is $6.47\times {10}^{-3}\text{J}$ .

(c)

To determine

The speed of the grape when it reaches the bottom of the bowl.

Answer to Problem 82PQ

The speed of the grape when it reaches the bottom of the bowl is $2.08\text{m}/\text{s}$ .

Explanation of Solution

Kinetic energy of the grape at bottom of bowl is obtained as $6.47\times {10}^{-3}\text{J}$.

Write the expression for kinetic energy of grape.

  $K_3=\frac{1}{2}m{v}_3^2$

Here, $m$ is the mass of grape and $v_3$ is the velocity of the grape at bottom of bowl.

Rearrange above equation to get $v_3$.

  $v_3=\sqrt{\frac{2K_3}{m}}$                                                                                                             (IV)

Conclusion:

Substitute $6.47\times {10}^{-3}\text{J}$ for $K_3$ and $3.00\text{g}$ for $m$ in equation (IV) to get $K_3$.

  $\begin{array}{c}{v}_3=\sqrt{\frac{2\left(6.47\times {10}^{-3}\text{J}\right)}{3.00\text{g}\times \frac{1\text{kg}}{1000\text{g}}}}\\ =2.08\text{m}/\text{s}\end{array}$

Therefore, the speed of the grape when it reaches the bottom of the bowl is $2.08\text{m}/\text{s}$ .

(d)

To determine

The potential and kinetic energies of the grape when it reaches a point that is height $h=15.0\text{cm}$ above the bottom of the bowl.

Answer to Problem 82PQ

The potential energy of the grape when it reaches a point that is height $h=15.0\text{cm}$ above the bottom of the bowl is $4.41\times {10}^{-3}\text{J}$ and the kinetic energy is $2.06\times {10}^{-3}\text{J}$ .

Explanation of Solution

Rewrite equation (I) to get potential energy at a height.

  $U=mgh$

Write conservation of energy equation as the grape moves from the top of the bowl to a height $h=15.0\text{cm}$ above the bottom of the bowl.

  $U_1+K_1=U_2+K_2$                                                                                                   (V)

Here, $U_2$ is the potential energy at height $h=15.0\text{cm}$ of bowl and $K_2$ is the kinetic energy at $h=15.0\text{cm}$ of bowl.

Conclusion:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $15.0\text{cm}$ for $h$ and $3.00\text{g}$ for $m$ equation (I) to get potential energy at $h=15.0\text{cm}$.

  $\begin{array}{c}{U}_2=\left(3.00\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(15.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =4.41\times {10}^{-3}\text{J}\end{array}$

In problem it is given that initially the ball is at rest at upper edge of bowl. Kinetic energy is zero at top edge of bowl. At top edge of bowl potential energy is $6.47\times {10}^{-3}\text{J}$.

Substitute $6.47\times {10}^{-3}\text{J}$ for $U_1$, $0\text{J}$ for $K_1$  and $4.41\times {10}^{-3}\text{J}$ for $U_2$ in equation (V) to get $K_2$.

  $\begin{array}{c}6.47\times {10}^{-3}\text{J}+0\text{J}=4.41\times {10}^{-3}\text{J}+K_2\\ K_2=6.47\times {10}^{-3}\text{J}-4.41\times {10}^{-3}\text{J}\\ =\text{2}\text{.06}\times {\text{10}}^{-3}\text{J}\end{array}$

Therefore, the potential energy of the grape when it reaches a point that is height $h=15.0\text{cm}$ above the bottom of the bowl is $4.41\times {10}^{-3}\text{J}$ and the kinetic energy is $2.06\times {10}^{-3}\text{J}$ .