# A model train, with a mass of 4 kg, is moving on a circular track with a radius of 15 m. If the train's kinetic energy changes from 32 j to 24 j, by how much will the centripetal force applied by the tracks change by?

Jul 11, 2017

The change in centripetal force is $= 1.07 N$

#### 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 radius is $= 15 m$

The variation of centripetal force is

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

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

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

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

$= \frac{2}{15} \cdot \left(32 - 24\right) N$

$= 1.07 N$

Jul 11, 2017

$\Delta {F}_{c} \approx - 1.0667 N$

#### Explanation:

$m = 4 k g$
$r = 15 m$
$K {E}_{i} = 32 J$
$K {E}_{f} = 24 J$

$\Delta K E = K {E}_{f} - K {E}_{i} = 24 - 32 = - 8 J$

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

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

$\Delta v = \sqrt{2 \frac{\Delta K E}{m}}$
$\Delta v = \sqrt{\frac{2 \cdot - 8}{4}}$
$\Delta v = \sqrt{- 4}$

${F}_{c} = m {a}_{c}$
where ${F}_{c}$ is centripetal force; ${a}_{c}$ is centripetal acceleration

${a}_{c} = \frac{{v}^{2}}{r}$
Thus,
${F}_{c} = m \frac{{v}^{2}}{r}$
$\Delta {F}_{c} = m \frac{{\left(\Delta v\right)}^{2}}{r}$
$\Delta {F}_{c} = 4 {\left(\sqrt{- 4}\right)}^{2} / 15$
$\Delta {F}_{c} = - \frac{16}{15} N$
$\Delta {F}_{c} \approx - 1.0667 N$

Note:
Negative sign signifies the centripetal force decreases