Question

A child’s pogo stick (Fig. P8.61) stores energy in a spring with a force constant of $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$ . At position A $\left(x_A=-100m\right)$, the spring compression is a maximum and the child is momentarily at rest. At position B $\left(x_B=0\right)$ the spring is relaxed and the child is moving upward. At position C, 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. (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.

Short answer

(a) The total energy of child-stick-earth system is 100 J.

(b) The position of child at point C is 0.410 m .

(c) The speed of the child at mean position is 2.84 m/s.

(d) The extension of spring at maximum kinetic energy of system is -9.8 mm.

(e) The maximum speed of child is 2.85 m/s.

Step-by-step solution

Identification of given data

The force constant of spring is $k=2.50\times {10}^4\mathrm{N}/\mathrm{m}$

The position of point A is $x_A=-0.100\mathrm{m}$

The position of point B is$x_B=0\mathrm{m}$

The combined mass of child and stick pogo is $m=25\mathrm{kg}$

The elastic potential energy and spring force is used to find the variables in the problem.

Determination of total energy of child-stick-Earth system

(a)

The total energy of child-stick-earth system is given as:

$E=mg{x}_A+\frac{1}{2}k{x}_A^2$

Substitute all the values in above equation.

$$\begin{gathered} E=\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(-0.100\mathrm{m}\right)+\frac{1}{2}\left(2.50\times {10}^4\mathrm{N}/\mathrm{m}\right){\left(-0.100\mathrm{m}\right)}^2 \\ \mathrm{E}=100.5\mathrm{J} \\ \mathrm{E}\approx 100\mathrm{J} \end{gathered}$$

Therefore, the total energy of child-stick-earth system is 100 J.

Determination of position of child at point C

(b)

The position of child at point C is given as:

$E=mg{x}_c$

Substitute all the values in above equation.

$$\begin{gathered} 100.5\mathrm{J}=\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)x_c \\ {x}_c=0.410\mathrm{m} \end{gathered}$$

Therefore, the position of child at point C is $0.410\mathrm{m}$.

Determination of speed of child at mean position

(c)

The speed of the child at mean position is given as:

$E=\frac{1}{2}m{v}^2$

Substitute all the values in above equation.

$$\begin{gathered} 100.5\mathrm{J}=\frac{1}{2}\left(25\mathrm{kg}\right)v^2 \\ v=2.84\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the child at mean position is $2.84\mathrm{m}/\mathrm{s}$.

Determination of extension of spring at maximum kinetic energy of system

(d)

The extension of spring at maximum kinetic energy of system is given as:

$$\begin{gathered} -kx=mg \\ x=-\frac{mg}{k} \end{gathered}$$

Substitute all the values in above equation.

$$\begin{gathered} x=-\frac{\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{2.50\times {10}^4\mathrm{N}/\mathrm{m}} \\ x=-98\times {10}^{-4}\mathrm{m} \\ \mathrm{x}=-\left(98\times {10}^4\mathrm{m}\right)\left(\frac{1000\mathrm{mm}}{1\mathrm{m}}\right) \\ \mathrm{x}=-9.8\mathrm{mm} \end{gathered}$$

Therefore, the extension of spring at maximum kinetic energy of system is -9.8 mm.

Determination of maximum speed of child

(e)

The maximum speed of child is given as:

$E=\frac{1}{2}m{v}_m^2+\frac{1}{2}k{x}^2+mgx$

Substitute all the values in above equation.

$$\begin{gathered} 100.5\mathrm{J}=\frac{1}{2}\left(25\mathrm{kg}\right)v_m^2+\frac{1}{2}\left(2.50\times {10}^4\mathrm{N}/\mathrm{m}\right){\left(-0.0098\mathrm{m}\right)}^2+\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(-0.0098\mathrm{m}\right) \\ {v}_m=2.85\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the maximum speed of child is 2.85 m/s.