# 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 0 J, by how much will the centripetal force applied by the tracks change by?

Feb 4, 2016

The centripetal force is given by $F = \frac{m {v}^{2}}{r}$. We are given the mass and the radius and can calculate the velocity from the kinetic energy. The change is $- 12$ $N$.

#### Explanation:

The centripetal force on an object of mass $m$ $k g$ in circular motion at velocity $v$ $m {s}^{-} 1$ in a circle of radius $r$ $m$ is given by:

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

In this case we know the mass is $3$ $k g$ and the radius is $3$ $m$, but the velocity is not immediately obvious. We are given the kinetic energy at two moments in time, though, and we know that kinetic energy is:

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

We can rearrange this to make $v$ the subject:

$v = \sqrt{\frac{2 {E}_{k}}{m}}$

Since the kinetic energy at the second time is $0$ $J$, the velocity is also $0$ $m {s}^{-} 1$, and therefore the centripetal force is also $0$ $N$.

At the first time, the kinetic energy is $18$ $J$, so:

$v = \sqrt{\frac{2 \cdot 18}{3}} = \sqrt{\frac{36}{3}} = \sqrt{12}$ $m {s}^{-} 1$

This means the centripetal force is:

$F = \frac{m {v}^{2}}{r} = \frac{3 \cdot 12}{3} = 12$ $N$ (since ${\left(\sqrt{12}\right)}^{2} = 12$)

So the centripetal force was $12$ $N$ at the beginning and $0$ $N$ at the end. Its change was therefore $- 12$ $N$.