Chapter 8, Problem 61AP

(a)

To determine

The total energy of the child stick Earth system .

Answer to Problem 61AP

The total energy of the child-stick Earth system is $100\text{J}$.

Explanation of Solution

Refer to Figure P8.61; Total energy of the system remains conserved at every point in the system.

Consider point A to calculate total energy. As the boy is momentarily at rest at point A Kinetic energy for the system is zero.

Write the expression for elastic potential energy stored in the spring.

  $U_s=\frac{1}{2}k{x}^2$                                                                                                     (I)

Here, $U_s$ is elastic potential energy, $k$ is force constant and $x$ is compression in the spring.

Write the expression for potential energy.

    $U_g=mgh$                                                                                                     (II)

Here, $U_g$ is gravitational potential energy, $m$ is mass, $g$ is acceleration due to gravity and $h$ is height.

Write the expression for kinetic energy of the system.

    $K=\frac{1}{2}m{v}^2$                                                                                                  (III)

Here, $K$ is kinetic energy and $v$ is velocity.

Write the expression for total energy at point A.

    $E_{\text{total}}=K_A+U_{gA}+U_{sA}$                                                                               (IV)

Here, $E_{\text{total}}$ is the total energy of the system, $K_A$ is kinetic energy at point A, $U_{gA}$ is gravitational potential energy at point A, $U_{sA}$ is the elastic potential energy at point A.

Substitute $\frac{1}{2}m{v}_A^2$ for $K_A$ , $\frac{1}{2}k{x}_A^2$ for $U_{sA}$ and $mg{x}_A$ for $U_{gA}$ in equation (IV)

    $E_{\text{total}}=\frac{1}{2}m{v}_A^2+mg{x}_A+\frac{1}{2}k{x}_A^2$

Here, $v_A$ is speed at A and $x_A$ is distance moved by the system at point A which is equal to compression of the string.

Substitute  $0$ for $\frac{1}{2}m{v}_A^2$ in above equation.   

    $E_{\text{total}}=0+mg{x}_A+\frac{1}{2}k{x}_A^2$                                                                            (V)

Conclusion:

Substitute  $2.50\times {10}^4\text{N}/\text{m}$ for $k$  , $-0.100\text{m}$ for $x_A$ , $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $25.0\text{kg}$ for $m$ in equation (V).

    $\begin{array}{c}{E}_{\text{total}}=\left[\begin{array}{c}0+\left(25.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.100\text{m}\right)\\ +\frac{1}{2}\left(2.50\times {10}^4\text{N}/\text{m}\right){\left(-0.100\text{m}\right)}^2\end{array}\right]\\ =-24.5+125\\ =100.5\text{J}\\ \simeq 100\text{J}\end{array}$

Thus, the total energy of the child-stick Earth system is $100\text{J}$.

(b)

To determine

The value of $x_c$.

Answer to Problem 61AP

The value of $x_c$ is $0.410\text{m}$ .

Explanation of Solution

Total energy at point A is equal to total energy at point C as the energy of system is conserved.

Write the expression for total energy at point C.

    $E_C=K_C+U_{gC}+U_{sC}$                                                                                   (VI)

Here, $E_C$ is the energy of the system at C, $K_C$ is kinetic energy at point C, $U_{gC}$ is gravitational potential energy at point C, and $U_{sC}$ is the elastic potential energy at point C.

Kinetic energy at C is zero as child again momentarily comes to rest at the top of the jump and elastic potential energy at C is also zero because string is no more compressed.

Substitute $\frac{1}{2}m{v}_C^2$ for $K_C$ , $\frac{1}{2}k{x}^2$ for $U_{sC}$ and $mg{x}_C$ for $U_{gC}$ in equation (VI).

    $E_C=\frac{1}{2}m{v}_C^2+mg{x}_C+\frac{1}{2}k{x}^2$

Here, $v_C$ is speed at C and $x_C$ is distance moved by the system at point C.

Substitute  $0$ for $\frac{1}{2}m{v}_C^2$ and $0$ for $\frac{1}{2}k{x}^2$ in above equation.   

    $E_C=0+mg{x}_C+0$                                                                                      (VII)

Write the expression for conservation of energy.

    $E_{\text{total}}=E_C$

Substitute $mg{x}_C$ for $E_C$ in above equation.

    $E_{\text{total}}=mg{x}_C$

Rearrange above equation for $x_c$ .

    $x_C=\frac{E_{\text{total}}}{mg}$                                                                                                (VIII)

Conclusion:

Substitute $100.5\text{J}$ for $E_{\text{total}}$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $25.0\text{kg}$ for $m$ in equation (VIII).

    $\begin{array}{c}{x}_C=\frac{\left(100.5\text{J}\right)}{\left(25.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}\\ =0.410\text{m}\end{array}$

Thus, the value of $x_c$ is $0.410\text{m}$ .

(c)

To determine

The speed of the child at point B.

Answer to Problem 61AP

Thus, the speed of the child at point B is $\text{2}\text{.84}\text{m/s}$.

Explanation of Solution

When $x=0$  i.e. at point B elastic potential and gravitational potential energy is zero.

Write the expression for total energy at point C.

    $E_B=K_B+U_{gB}+U_{sB}$                                                                                   (IX)

Here, $E_B$ is the energy of the system at B, $K_B$ is kinetic energy at point B, $U_{gB}$ is gravitational potential energy at point B, and $U_{sB}$ is the elastic potential energy at point B.

Substitute $\frac{1}{2}m{v}_B^2$ for $K_B$ , $\frac{1}{2}k{x}^2$ for $U_{sB}$ and $mg{x}_B$ for $U_{gB}$ in equation (IX).

    $E_B=\frac{1}{2}m{v}_B^2+mg{x}_B+\frac{1}{2}k{x}^2$

Here, $v_B$ is speed at C and $x_B$ is distance moved by the system at point B.

Substitute  $0$ for $mg{x}_B$ and $0$ for $\frac{1}{2}k{x}^2$ in above equation.   

    $E_B=\frac{1}{2}m{v}_B^2+0+0$                                                                                       (X)

Write the expression for conservation of energy.

    $E_{\text{total}}=E_B$

Substitute $\frac{1}{2}m{v}_B^2$ for $E_B$ in above equation.

    $E_{\text{total}}=\frac{1}{2}m{v}_B^2$

Rearrange above equation for $v_B$ .

    $v_B=\sqrt{\frac{2E_{total}}{m}}$                                                                                               (XI)

Conclusion:

Substitute $100.5\text{J}$ for $E_{total}$  and $25.0\text{kg}$ for $m$ in equation (XI).

    $\begin{array}{c}{v}_B=\sqrt{\frac{2\left(100.5\text{J}\right)}{\left(25.0\text{kg}\right)}}\\ =2.835\text{m/s}\\ \simeq \text{2}\text{.84}\text{m/s}\end{array}$

Thus, the speed of the child at point B is $\text{2}\text{.84}\text{m/s}$.

(d)

To determine

The value of $x$ for which the kinetic energy of the system is a maximum.

Answer to Problem 61AP

The value of $x$ for which the kinetic energy of the system is a maximum is $-9.80\text{mm}$.

Explanation of Solution

To calculate the value of $x$ for which the kinetic energy of the system is maximum..

Write the expression for total energy at any point.

    $E=K+U_g+U_s$

Here, $E$ is the total energy of the system .

Rearrange above equation for $K$ .

    $K=E-U_g-U_s$

Substitute $\frac{1}{2}k{x}^2$ for $U_s$ and $mgx$ for $U_g$ in above equation.

    $K=E-mgx-\frac{1}{2}k{x}^2$

Substitute  $\left(-x\right)$ for $x$ in above equation as     $x$ is the compression of the spring its movement is in opposite direction.

    $K=E-mg\left(-x\right)-\frac{1}{2}k{\left(-x\right)}^2$

Simplify above expression.

  $K=E+mgx-\frac{1}{2}k{\left(x\right)}^2$                                                                              (XII)   

Differentiate the equation with respect to $x$ and equate the result to zero

Differentiate equation (XII) with respect to $x$ .

    $\frac{dK}{dx}=0+mg-kx$

Equate it to zero.

    $0+mg-kx=0$

Rearrange above equation for $x$ .

    $x=\frac{mg}{k}$                                                                                                     (XIII)

Conclusion:

Substitute $2.50\times {10}^4\text{N}/\text{m}$ for $k$ , $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $25.0\text{kg}$ for $m$ in equation (XIII).

    $\begin{array}{c}x=\frac{\left(25.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)}{\left(2.50\times {10}^4\text{N}/\text{m}\right)}\\ \text{=}0.0098\text{m}\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =9.80\text{mm}\end{array}$

As $x$ is value of compression and it is below $x=0$ value of $x$ is negative. Therefore value of $x$ is $-9.80\text{mm}$.

Thus, the value of $x$ for which the kinetic energy of the system is a maximum is $-9.80\text{mm}$.

(e)

To determine

The child’s maximum upward speed.

Answer to Problem 61AP

The child’s maximum upward speed is $2.85\text{m/s}$.

Explanation of Solution

Consider point D where velocity is maximum and it is the point where we have maximum kinetic energy.

Write the expression for total energy at point D.

    $E_D=K_D+U_{gD}+U_{sD}$                                                                               (XIV)

Here, $E_D$ is the energy of the system at D, $K_D$ is kinetic energy at point D, $U_{gD}$ is gravitational potential energy at point D, and $U_{sD}$ is the elastic potential energy at point D.

Write the expression for conservation of energy between point A and D.

    $E_{\text{total}}=E_D$

Substitute $K_D+U_{gD}+U_{sD}$ for $E_D$ in above equation.

    $E_{\text{total}}=K_D+U_{gD}+U_{sD}$

Rearrange above equation for $K_D$ .

  $K_D=E_{\text{total}}-U_{gD}-U_{sD}$                                                                           (XV)

Substitute $\frac{1}{2}m{v}_D^2$ for $K_D$ , $\frac{1}{2}k{x}_D^2$ for $U_{sD}$ and $mg{x}_D$ for $U_{gD}$ in equation (XV)

    $\frac{1}{2}m{v}_D^2=E_{\text{total}}-mg{x}_D-\frac{1}{2}k{x}_D^2$

Here, $v_D$ is speed at D and $x_D$ is distance moved by the system at point D.

Multiply above equation by $2$ .

    $m{v}_D^2=2E_{\text{total}}-2mg{x}_D-k{x}_D^2$

Rearrange above equation for $v_D$

    $v_D=\sqrt{\frac{2E_{\text{total}}-2mg{x}_D-k{x}_D^2}{m}}$                                                                 (XVI)

Conclusion:

Substitute $2.50\times {10}^4\text{N}/\text{m}$ for $k$, $-9.80\text{mm}$ for $x_D$ , $100.5\text{J}$ for $E_{\text{total}}$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $25.0\text{kg}$ for $m$ in equation (XVI).

    $\begin{array}{c}{v}_D=\sqrt{\frac{\left[\begin{array}{l}2\left(100.5\text{J}\right)-2\left(25.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(-9.80\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\\ -\left(2.50\times {10}^4\text{N}/\text{m}\right){\left(\left(-9.80\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\end{array}\right]}{\left(25.0\text{kg}\right)}}\\ =\sqrt{\frac{201\text{J+4}\text{.802}\text{J -2}\text{.401}\text{J}}{\left(25.0\text{kg}\right)}}\\ =2.85\text{m/s}\end{array}$

Thus, the child’s maximum upward speed is $2.85\text{m/s}$.