A model train, with a mass of 8 kg, is moving on a circular track with a radius of 2 m. If the train's kinetic energy changes from 8 j to 36 j, by how much will the centripetal force applied by the tracks change by?

Nov 9, 2017

The centripetal force increases by 28 N, from 8 N to 36 N.
Details follow.

Explanation:

Centripetal force is given by

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

For this train, we know the mass of the train, $m$ and the radius $r$ of the circular path it is following. What we need is the speed of the train $v$.

Since the kinetic energy $K$ is given, we can use this to calculate the speed:

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

which I will rewrite as

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

(because we will need ${v}^{2}$ for the centripetal force formula, rather than $v$).

When the energy is 8 J, ${v}^{2} = \frac{2 \cdot 8}{8} = 2$

and the centripetal force is ${F}_{c} = \frac{m {v}^{2}}{r} = \frac{8 \cdot 2}{2} = 8 N$

When the energy is 36 J, ${v}^{2} = \frac{2 \cdot 36}{8} = 9$

and the centripetal force is ${F}_{c} = \frac{m {v}^{2}}{r} = \frac{8 \cdot 9}{2} = 36 N$

The change in centripetal force will be $36 - 8 = 28 N$