Suggested languages for you:

Americas

Europe

Q22 PE

Expert-verifiedFound in: Page 261

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{-}}{\mathbf{\text{kg}}}$**** subway train is brought to a stop from a speed of ${\mathbf{0}}{\mathbf{.}}{\mathbf{500}}{\mathbf{\text{m}}}{\mathbf{/}}{\mathbf{\text{s}}}$**** in ${\mathbf{0}}{\mathbf{.}}{\mathbf{400}}{\mathbf{\text{m}}}$**** by a large spring bumper at the end of its track. What is the force constant k of the spring?**

The force constant of the spring is $781250\text{N}/\text{m}$.

**Force constant: When a spring is stretched by the application of some force, then according to the Hooke’s law the increase in the length of spring from its equilibrium length is directly proportional to the force applied.**

Mathematically,

$F\propto x\phantom{\rule{0ex}{0ex}}F=kx$

Here, k is the proportionality constant known as force constant or spring constant.

** **

When a spring is stretched, the change in potential energy of the spring is,

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

Here, k is the force constant, and x is the displacement $\left(x=0.400\text{m}\right)$.

According to the law of conservation of energy,

$\Delta \text{PE}=\Delta \text{KE}\phantom{\rule{0ex}{0ex}}\frac{1}{2}k{x}^{2}=\frac{1}{2}m{v}^{2}$

Here, m is the mass of the subway train $\left(m=5.00\times {10}^{5}\text{kg}\right)$, and v is the initial velocity of the subway train $\left(v=0.500\text{m}/\text{s}\right)$.

The expression for the force constant is,

$k=\frac{m{v}^{2}}{{x}^{2}}$

Putting all known values,

$k=\frac{\left(5.00\times {10}^{5}\text{kg}\right)\times {\left(0.500\text{m}/\text{s}\right)}^{2}}{{\left(0.400\text{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=781250\text{N}/\text{m}$

Therefore, the required force constant of the spring is $781250\text{N}/\text{m}$.

94% of StudySmarter users get better grades.

Sign up for free