# A model train, with a mass of 8 kg, is moving on a circular track with a radius of 1 m. If the train's kinetic energy changes from 32 J to 0 J, by how much will the centripetal force applied by the tracks change?

Jun 24, 2016

The centripetal force is given by $F = \frac{m {v}^{2}}{r}$, and we can calculate the change in $v$ from the change in the kinetic energy, ${E}_{k} = \frac{1}{2} m {v}^{2}$. The change in the centripetal force acting is $- 64$ $N$.

#### Explanation:

The kinetic energy is given by ${E}_{k} = \frac{1}{2} m {v}^{2}$, and we can rearrange to make $v$ the subject:

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

The final kinetic energy is zero and therefore the final velocity is zero.

We can use the mass, radius and initial kinetic energy to calculate the initial velocity:

$v = \sqrt{\frac{2 {E}_{k}}{m}} = \sqrt{\frac{2 \cdot 32}{8}} = \sqrt{\frac{64}{8}} = \sqrt{8} \approx 2.8$ $m {s}^{-} 1$

Since the final velocity is zero, the final centripetal force will be zero. The initial centripetal force will be:

$F = \frac{m {v}^{2}}{r} = \frac{8 \cdot {2.8}^{2}}{1} = 64$ $N$.

Since the final centripetal force is $0$ $N$, the change in the centripetal force is:

final - initial = $0 - 64 = - 64$ $N$