# A model train, with a mass of 3 kg, is moving on a circular track with a radius of 3 m. If the train's kinetic energy changes from 18 j to 27 j, by how much will the centripetal force applied by the tracks change by?

May 9, 2018

The change in centripetal force is $= 6 N$

#### Explanation:

The mass of the train is $m = 3 k g$

The radius of the track is $r = 3 m$

The centripetal force is

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

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

The kinetic energy is

$K E = \frac{1}{2} m {v}^{2}$

Therefore,

The variation of kinetic energy is

$\Delta K E = \frac{1}{2} m \left({v}_{1}^{2} - {v}_{2}^{2}\right)$

${v}_{1}^{2} = \frac{{F}_{1} r}{m}$

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

So,

$\Delta K E = \frac{1}{2} m \left(\frac{{F}_{1} r}{m} - \frac{{F}_{2} r}{m}\right) = \frac{r}{2} \left({F}_{2} - {F}_{1}\right) = \frac{r}{2} \Delta F$

$\Delta F = \frac{2}{r} \Delta K E$

The change in centripetal force is

$\Delta F = \frac{2}{3} \cdot \left(27 - 18\right) = 6 N$