A model train, with a mass of 4 kg, is moving on a circular track with a radius of 3 m. If the train's kinetic energy changes from 12 j to 21 j, by how much will the centripetal force applied by the tracks change by?

Apr 5, 2017

The change in centripetal force is $= 6 N$

Explanation:

The centripetal force is

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

The kinetic energy is

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

The variation of kinetic energy is

$\Delta K E = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2}$

$= \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

The variation of centripetal force is

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

$\Delta F = 2 \frac{m}{r} \frac{1}{2} \left({v}^{2} - {u}^{2}\right)$

$= \frac{2}{r} \cdot \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$= \frac{2}{r} \cdot \Delta K E$

$= \frac{2}{3} \cdot \left(21 - 12\right) N$

$= 6 N$

Apr 5, 2017

The change in centripetal force is $= 6 N$

Explanation:

The centripetal force is

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

The kinetic energy is

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

The variation of kinetic energy is

$\Delta K E = \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2}$

$= \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

The variation of centripetal force is

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

$\Delta F = 2 \frac{m}{r} \frac{1}{2} \left({v}^{2} - {u}^{2}\right)$

$= \frac{2}{r} \cdot \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$= \frac{2}{r} \cdot \Delta K E$

$= \frac{2}{3} \cdot \left(21 - 12\right) N$

$= 6 N$