# A model train, with a mass of 2 kg, is moving on a circular track with a radius of 6 m. If the train's kinetic energy changes from 2 j to 96 j, by how much will the centripetal force applied by the tracks change by?

Feb 18, 2016

The centripetal force increases by $31 \frac{1}{3} N$

#### Explanation:

This is the type of question that seems complex when you read it, but it is simple when you break it down into small steps. This because the inputs (Kinetic Energy) seem to have no relation so the outputs (centripetal force). There must be a link which we can find by writing the equations for each down to begin with:

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

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

From this we notice that $v$ is the common element. If we find the change in $v$ due to the kinetic energy change, we can find the corresponding change in centripetal force.

$K . E {.}_{1} = \frac{1}{2} m {v}_{1}^{2}$ where $K . E {.}_{1} = 2 J$

$2 J = \frac{1}{2} 2 k g \cdot {v}_{1}^{2}$

${v}_{1} = \sqrt{2} m / s$

and

$K . E {.}_{2} = \frac{1}{2} m {v}_{2}^{2}$ where $K . E {.}_{1} = 96 J$

${v}_{2} = \sqrt{96} m / s$

Therefore, we can find the centripetal force in the beginning and end states from:

${F}_{c 1} = \frac{m {v}_{1}^{2}}{r} = \frac{2 k g \cdot 2 {m}^{2} / {s}^{2}}{6 m} = \frac{2}{3} N$

and

${F}_{c 2} = \frac{m {v}_{2}^{2}}{r} = \frac{2 k g \cdot 96 {m}^{2} / {s}^{2}}{6 m} = 32 N$

so the change in centripetal force is the difference of these two forces:

$\Delta {F}_{c} = {F}_{c 2} - {F}_{c 1} = 32 N - \frac{2}{3} N = 31 \frac{1}{3} N$