# A model train, with a mass of 15 kg, is moving on a circular track with a radius of 6 m. If the train's kinetic energy changes from 40 j to 27 j, by how much will the centripetal force applied by the tracks change by?

Jan 27, 2018

$- 4.3$ Newtons.

#### Explanation:

We know that the centripetal force is given by

${F}_{c} = \frac{m {v}^{2}}{r}$

We also know that kinetic energy is $E = \frac{1}{2} m {v}^{2}$. We know that $2 E = m {v}^{2}$, or ${v}^{2} = \frac{2 E}{m}$. Therefore:

${F}_{c} = \frac{m \frac{2 E}{m}}{r}$

${F}_{c} = \frac{2 E}{r}$

So

$\Delta {F}_{c} = \frac{\Delta 2 E}{r}$

Or

$\Delta {F}_{c} = \frac{2 \left(27\right)}{6} - \frac{2 \left(40\right)}{6}$

$\Delta {F}_{c} = 9 - \frac{40}{3}$

$\Delta {F}_{c} = - 4.3 N$

Hopefully this helps!