Suggested languages for you:

Americas

Europe

Q23 PE

Expert-verifiedFound in: Page 261

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A pogo stick has a spring with a force constant of ${\mathbf{2}}{\mathbf{.}}{\mathbf{50}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{4}}}{\mathbf{\text{N}}}{\mathbf{/}}{\mathbf{\text{m}}}$****, which can be compressed ${\mathbf{12}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{cm}}}$****. To what maximum height can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of ${\mathbf{40}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{kg}}}$****? Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy.**

The child can jump up to the maximum height of $0.46\text{m}$.

**Spring potential energy: When a spring is stretched or compressed by the application of force, the potential energy stored in the spring is,**

${\text{PE}}_{s}=\frac{1}{2}k{x}^{2}$

Here, k is the force constant of the spring $\left(k=2.50\times {10}^{4}\text{N}/\text{m}\right)$ and x is the compression in the spring 12.0 cm.

When a child jumps on the spring, the spring potential energy in the spring gets converted into gravitational potential energy and the child attains some height. Mathematically,

${\text{PE}}_{s}={\text{PE}}_{g}\phantom{\rule{0ex}{0ex}}\frac{1}{2}k{x}^{2}=mgh$

Here, m is the mass of the child $\left(m=40.0\text{kg}\right)$, g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$, and *h* is the maximum height attained by the child.

The expression for the maximum height attained by the child is,

$h=\frac{k{x}^{2}}{2mg}$

Putting all known values,

$h=\frac{\left(2.50\times {10}^{4}\text{N}/\text{m}\right)\times {\left(12.0\text{cm}\right)}^{2}}{\left(40.0\text{kg}\right)\times \left(9.8\text{m}/{\text{s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(2.50\times {10}^{4}\text{N}/\text{m}\right)\times {\left[\left(12.0\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right]}^{2}}{2\times \left(40.0\text{kg}\right)\times \left(9.8\text{m}/{\text{s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=0.46\text{m}$

Therefore, the child can jump up to the maximum height of $0.46\text{m}$.

94% of StudySmarter users get better grades.

Sign up for free