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Q23 PE

Expert-verified
Found in: Page 261

### College Physics (Urone)

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 ${\mathbf{2}}{\mathbf{.}}{\mathbf{50}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{4}}}{\mathbf{\text{N}}}{\mathbf{/}}{\mathbf{\text{m}}}$, 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 $0.46\text{m}$.

See the step by step solution

## Spring Potential Energy

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

## Find the maximum height can a child jump

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,

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

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