# A model train, with a mass of 4 kg, is moving on a circular track with a radius of 3 m. If the train's kinetic energy changes from 8 j to 48 j, by how much will the centripetal force applied by the tracks change by?

Mar 6, 2016

$\delta F = 26.6666666666 \ldots \ldots . N$

#### Explanation:

Given we have to find the centripetal force on a model train of mass $m = 4 k g$ which is moving in a circular track of radius $r = 3 m$

The centripetal force of a body is $F = m {v}^{2} / r$ and the kinetic energy of a body is $K = \frac{1}{2} m {v}^{2} \setminus \implies 2 k = m {v}^{2}$
Substituting $K$ in $F$, we get $F = 2 \frac{K}{r}$ or $\delta F = \frac{2}{r} \setminus \delta K$

Here, $\delta K = 48 - 8 = 40 J$
So, change in centripetal force applied by tracks change by $\delta F = 2 \cdot \frac{40}{3}$

Um.....yeah, that's why you end up with $666666$s