# A model train, with a mass of 6 kg, is moving on a circular track with a radius of 4 m. If the train's kinetic energy changes from 24 j to 48 j, by how much will the centripetal force applied by the tracks change by?

Mar 4, 2017

$2.05 N$

#### Explanation:

The kinetic energy of the train is given by

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

which we can rearrange to find the velocity (which we need) as

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

so, calculating the velocity in both cases ($24 J$ and $48 J$),

${v}_{1} = \sqrt{\frac{2 \times 24}{6}} = \sqrt{8} \approx 2.83 m {s}^{-} 1$

${v}_{2} = \sqrt{\frac{2 \cdot 48}{6}} = \sqrt{16} = 4 m {s}^{-} 1$

Therefore,

$\Delta v$ = ${v}_{2} - {v}_{2} = 4 - 2.83 = 1.17 m {s}^{-} 1$

The equation of centripetal force is

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

where $F$ is force, $m$ is mass, ${a}_{c}$ is centripetal acceleration, $v$ is velocity, and $r$ is the radius of the circle.

Therefore we can say that change in centripetal force is

$\Delta F = \frac{m {\left(\Delta v\right)}^{2}}{r}$

which, putting in the values we know, is

$\Delta F = \frac{6 \times {1.17}^{2}}{4} \approx 2.05 N$