Chapter 7, Problem 69P

A child’s pogo stick (Fig. P7.69) stores energy in a spring with a force constant of 2.50 × 10⁴ N/m. At position Ⓐ (x_Ⓐ = −0.100 m), the spring compression is a maximum and the child is momentarily at rest. At position Ⓑ (x_Ⓑ = 0), the spring is relaxed and the child is moving upward. At position Ⓒ, the child is again momentarily at rest at the top of the jump. The combined mass of child and pogo stick is 25.0 kg. Although the boy must lean forward to remain balanced, the angle is small, so let’s assume the pogo stick is vertical. Also assume the boy does not bend his legs during the motion. (a) Calculate the total energy of the child–stick–Earth system, taking both gravitational and elastic potential energies as zero for x = 0. (b) Determine x_Ⓒ. (c) Calculate the speed of the child at x = 0. (d) Determine the value of x for which the kinetic energy of the system is a maximum. (e) Calculate the child’s maximum upward speed.

Figure P7.69

To determine

Total energy of the system.

Answer to Problem 69P

Total energy is $\underset{\_}{100\text{J}}$.

Explanation of Solution

Write the equation for total energy of the system

    $E_{\text{mech}}=\Delta K+\Delta U+\Delta U_s$        (I)

Here $E_{\text{mech}}$ is the total energy of the system, $\Delta K$ is the change in kinetic energy, $\Delta U$ is the gravitational potential energy and $\Delta U_s$ is the change in elastic potential energy.

Write the equation for kinetic energy at A, B and C

  $\begin{array}{c}{K}_A=\frac{1}{2}m{v}_A{}^2\\ K_B=\frac{1}{2}m{v}_B{}^2\\ K_C=\frac{1}{2}m{v}_C{}^2\end{array}$        (II)

Here $K_A$ is the kinetic energy at A, $K_B$ is the kinetic energy at B, $K_C$ is the kinetic energy at C $m$ is the mass of the block, $v_A$ is the velocity at point A, $v_B$ is the velocity at point B and $v_C$ is the kinetic energy at point C.

Write the expression for gravitational potential energy at A, B and C.

  $\begin{array}{c}{U}_A=mg{x}_A\\ U_B=mg{x}_B\\ U_C=mg{x}_C\end{array}$        (III)

Here $U_A$ is the gravitational potential energy at point A, $U_B$ is the gravitational potential energy at point B, $U_C$ is the gravitational potential energy at point C, $m$ is the mass of the block, $g$ is the acceleration due to gravity, $x_A$ is the extension at A, $x_B$ is the extension at B and $x_C$ is the extension at C.

Write down the equation for elastic potential energy at A, B and C.

  $\begin{array}{l}{U}_{sA}=\frac{1}{2}k{x}_A{}^2\\ U_{sB}=\frac{1}{2}k{x}_B{}^2\\ U_{sC}=\frac{1}{2}k{x}_C{}^2\end{array}$        (IV)

Here $U_{sA}$, $U_{sB}$ and $U_{sC}$ is the elastic potential energy at point A, B and C respectively and $k$ is the spring constant

As there is no motion, kinetic energy is zero

Substitute $U_A$ and $U_{sA}$ in (I)

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

Conclusion:

Substitute $25\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $-0.100\text{m}$ for $x_A$ and $2.5\times {10}^4\text{N}/\text{m}$ for $k$ in (V)

    $\begin{array}{c}{E}_{\text{mech}}=\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.100\text{m}\right)+\frac{1}{2}\left(2.5\times {10}^4\text{N}/\text{m}\right){\left(-0.100\text{m}\right)}^2\\ =100\text{J}\end{array}$

Total energy is $\underset{\_}{100\text{J}}$.

To determine

Extension at point C.

Answer to Problem 69P

Extension is $\underset{\_}{0.410\text{m}}$.

Explanation of Solution

Write the energy conservation equation

    $K_C+U_C+U_{sC}=K_A+U_A+U_{sA}$        (VI)

Substitute for all the terms in (VI)

    $\frac{1}{2}m{v}_C{}^2+mg{x}_C+\frac{1}{2}k{x}_C{}^2=\frac{1}{2}m{v}_A{}^2+mg{x}_A+\frac{1}{2}k{x}_A{}^2$        (VII)

Kinetic energy is zero as there is no motion

The elastic potential energy is zero at C.

Now rewrite (VII) in terms of $x_C$

    $x_C=\frac{mg{x}_A+\frac{1}{2}k{x}_A{}^2}{mg}$        (VIII)

Conclusion:

Substitute $25\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $-0.100\text{m}$ for $x_A$ and $2.5\times {10}^4\text{N}/\text{m}$ for $k$ in (VIII)

    $\begin{array}{c}{x}_C=\frac{\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.100\text{m}\right)+\frac{1}{2}\left(2.5\times {10}^4\text{N}/\text{m}\right){\left(-0.100\text{m}\right)}^2}{\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ =0.410\text{m}\end{array}$

Extension is $\underset{\_}{0.410\text{m}}$.

To determine

Speed of the child at $x=0$

Answer to Problem 69P

The speed is $\underset{\_}{2.\text{84}\text{m}/\text{s}}$.

Explanation of Solution

Write the energy conservation equation.

    $K_B+U_B+U_{sB}=K_A+U_A+U_{sA}$        (IX)

At $x=0$, $U_B=U_{sB}=0$

Kinetic energy at A is also zero.

Then rewrite (IX) in terms of $v_b$

    $\begin{array}{c}\frac{1}{2}m{v}_B{}^2=mg{x}_A+\frac{1}{2}k{x}_A{}^2\\ v_B=2\left(\frac{mg{x}_A+\frac{1}{2}k{x}_A{}^2}{m}\right)\end{array}$        (X)

Conclusion:

Substitute $25\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $-0.100\text{m}$ for $x_A$ and $2.5\times {10}^4\text{N}/\text{m}$ for $k$ in (X)

    $\begin{array}{c}{v}_B=2\left(\frac{\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.100\text{m}\right)+\frac{1}{2}\left(2.5\times {10}^4\text{N}/\text{m}\right){\left(-0.100\text{m}\right)}^2}{25\text{kg}}\right)\\ =2.84\text{m}/\text{s}\end{array}$

The speed is $\underset{\_}{2.\text{84}\text{m}/\text{s}}$.

To determine

Value of $x$ where kinetic energy is maximum.

Answer to Problem 69P

Kinetic energy is maximum at $x=-\underset{\_}{9.8\text{mm}}$.

Explanation of Solution

Write the energy of the system when the spring is compressed

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

Rewrite (XI) in terms of $K$ and differentiate with respect to $x$.

  $\begin{array}{c}\frac{dK}{dx}=0\\ 0=\frac{d\left(E-\frac{1}{2}k{x}^2+mgx\right)}{dx}\\ 0=-kx+mg\end{array}$        (XII)

Rewrite (XII) in terms of $x$.

    $x=\frac{mg}{k}$        (XIII)

Conclusion:

Substitute $25\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2.5\times {10}^4\text{N}/\text{m}$ for $k$ in (XIII)

    $\begin{array}{c}x=\frac{\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(2.5\times {10}^4\text{N}/\text{m}\right)}\\ =0.0098\text{m}\\ \text{=0}\text{.98}\text{mm}\end{array}$

As this the value for compression, the position is below $x=0$

Therefore,

Kinetic energy is maximum at $x=-\underset{\_}{9.8\text{mm}}$.

To determine

Maximum upward speed.

Answer to Problem 69P

Maximum speed is $\underset{\_}{2.85\text{m}/\text{s}}$.

Explanation of Solution

Write the equation for maximum kinetic energy

    $K_{\max }=K_A+\left(U_A-{U|}_{x=-9.8\text{mm}}\right)+\left(U_{sA}-{U_s|}_{x=-9.8\text{mm}}\right)$        (XIV)

$K_A=0$

Substitute  (III) and (IV) in (XIV)

    $\frac{1}{2}m{v}^2{}_{\max }=\left(mg{x}_{-0.100}-mg{x}_{-9.8\text{mm}}\right)+\left(\frac{1}{2}k{x}^2{}_{-0.100}-\frac{1}{2}k{x}^2{}_{-9.8\text{mm}}\right)$        (XV)

Rewrite (XV) in terms of $v_{\max }$.

    $v_{\max }=\sqrt{2\frac{\left(\left(mg{x}_{-0.100}-mg{x}_{-9.8\text{mm}}\right)+\left(\frac{1}{2}k{x}^2{}_{-0.100}-\frac{1}{2}k{x}^2{}_{-9.8\text{mm}}\right)\right)}{m}}$        (XVI)

Conclusion:

Substitute $25\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $2.5\times {10}^4\text{N}/\text{m}$ for $k$ in (XIV)

    $\begin{array}{c}{v}_{\max }=\sqrt{2\frac{\left(\left(\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.100\text{m}\right)-\left(25\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-0.0098\text{m}\right)\right)+\left(\frac{1}{2}\left(2.5\times {10}^4\text{N}/\text{m}\right){\left(-0.100\text{m}\right)}^2-\frac{1}{2}\left(2.5\times {10}^4\text{N}/\text{m}\right){\left(\left(-0.0098\text{m}\right)\right)}^2\right)\right)}{\left(25\text{kg}\right)}}\\ =2.85\text{N}/\text{m}\end{array}$

Maximum speed is $\underset{\_}{2.85\text{m}/\text{s}}$.