Chapter 7, Problem 45P

A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ.

Figure P7.45 Problems 45 and 46.

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is $\underset{\_}{0.588\text{J}}$.

Explanation of Solution

Write down the equation for gravitational potential energy

  $U_B=mgR$        (I)

Here $U_B$ is the gravitational potential energy at point B, $m$ is the mass, $g$ is the acceleration due to gravity and $R$ is the distance.

Conclusion:

Substitute $200\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30\text{cm}$ for $R$ in (I).

  $\begin{array}{c}U=\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(30\times {10}^{-2}\text{m}\right)\\ =0.588\text{J}\end{array}$

The gravitational potential energy is $\underset{\_}{0.588\text{J}}$.

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is $\underset{\_}{0.588\text{J}}$.

Explanation of Solution

Write the law of conservation of energy.

  $K_A+U_A=K_B+U_B$        (II)

Here $K_A$ is the kinetic energy at point A, $U_A$ is the potential energy at point A, $K_B$ is the kinetic energy at point B and $U_B$ is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

  $K_B=U_A$        (III)

Substitute (I) in (III)

  $K_B=mgR$        (IV)

Conclusion:

Substitute $200\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30\text{cm}$ for $h$ in (I).

  $\begin{array}{c}U=\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(30\times {10}^{-2}\text{m}\right)\\ =0.588\text{J}\end{array}$

The gravitational potential energy is $\underset{\_}{0.588\text{J}}$.

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is $\underset{\_}{2.42\text{m}/\text{s}}$

Explanation of Solution

Write down the equation for kinetic energy.

  $K_B=\frac{1}{2}m{v}_B{}^2$        (V)

Here $K_B$ is the kinetic energy at point B, $m$ is the mass of the block, $v_B$ is the velocity

Rewrite (V) in terms of $v_B$.

  $v_B=\sqrt{\frac{2K_B}{m}}$        (VI)

Conclusion:

Substitute $0.588\text{J}$ for $K_B$ and $200\text{g}$ for $m$ in (VI).

  $\begin{array}{c}{v}_B=\sqrt{\frac{2\left(0.588\text{J}\right)}{\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)}}\\ =\sqrt{\frac{2\left(0.588\text{J}\right)}{\left(200\times {10}^{-3}\text{kg}\right)}}\\ =2.42\text{m}/\text{s}\end{array}$

Velocity is $\underset{\_}{2.42\text{m}/\text{s}}$

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is $\underset{\_}{0.392\text{J}}$.

Kinetic energy is $\underset{\_}{0.196\text{J}}$

Explanation of Solution

Write down the equation for gravitational potential energy

  $U_C=mg{h}_C$        (VII)

Here $U_C$ is the gravitational potential energy at point B, $m$ is the mass, $g$ is the acceleration due to gravity and $h_C$ is the height at point C.

Write the law of conservation of energy.

  $K_A+U_A=K_C+U_C$        (VIII)

Here $K_A$ is the kinetic energy at point A, $U_A$ is the potential energy at point A, $K_C$ is the kinetic energy at point C and $U_C$ is the potential energy at point C.

Rewrite (VIII)

  $K_C=K_A+U_A-U_C$

At point A, kinetic energy is zero.

Therefore,

    $K_C=U_A-U_C$        (IX)

Conclusion:

Substitute $200\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $20\text{cm}$ for $h_c$ in (VII).

  $\begin{array}{c}U=\left(200\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(20\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(20\times {10}^{-2}\text{m}\right)\\ =0.392\text{J}\end{array}$

Substitute $0.588\text{J}$ for $U_A$ and $0.392\text{J}$ for $U_C$ in (IX)

    $\begin{array}{c}{K}_C=0.588\text{J}-0.392\text{J}\\ \text{=0}\text{.196}\text{J}\end{array}$

Potential energy is $\underset{\_}{0.392\text{J}}$.

Kinetic energy is $\underset{\_}{0.196\text{J}}$