# A model train, with a mass of 2 kg, is moving on a circular track with a radius of 3 m. If the train's kinetic energy changes from 3 j to 5 j, by how much will the centripetal force applied by the tracks change by?

Feb 15, 2016

$= 1. \dot{3} N$

#### Explanation:

For Circular Motion, Centripetal Force ${F}_{c} = - m r {\omega}^{2}$

Where $m$ is the mass of the body in circular motion. $r$ is the radius of the circle and $\omega$ is angular velocity.
$-$ sign means that the force is opposite to the radius vector and is directed towards the center.
Now velocity $v = r \omega$
Therefore, magnitude of the force $| {F}_{c} | = \frac{m {v}^{2}}{r}$
Kinetic energy is given by
$K E = \frac{1}{2} m {v}^{2}$
Let the velocity of the train change from ${v}_{i}$ to ${v}_{f}$
Given is change in kinetic energy $\Delta K E = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$
$\frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2} = 5 - 3 = 2$
or $\frac{m}{2} \left({v}_{f}^{2} - {v}_{i}^{2}\right) = 2$
or $m \left({v}_{f}^{2} - {v}_{i}^{2}\right) = 4$ ........(1)

Change in the centripetal force applied by the tracks $= \frac{m {v}_{f}^{2}}{r} - \frac{m {v}_{i}^{2}}{r}$
$= \frac{m}{r} \left({v}_{f}^{2} - {v}_{i}^{2}\right)$ ........(2)
Inserting value of $m \left({v}_{f}^{2} - {v}_{i}^{2}\right)$ from (1) and given value of $r$

Change in the centripetal force applied by the tracks $= \frac{1}{3} \cdot 4$
$= 1. \dot{3} N$