# A model train, with a mass of 6 kg, is moving on a circular track with a radius of 1 m. If the train's kinetic energy changes from 27 j to 36 j, by how much will the centripetal force applied by the tracks change by?

Jan 13, 2018

18 N

#### Explanation:

Let'assume Kinetic energy in the 1st case is K1 and in 2nd case K2
So,$\frac{K 2}{K 1}$ = $\left(\frac{m}{m}\right) \left({\left(v 2\right)}^{2} / {\left(v 1\right)}^{2}\right)$ so, ratio of square of v2 and v1 will be $\frac{36}{27}$
Now, Centripetal force F is given as $m \frac{{v}^{2}}{r}$
As,mass and radius are unchanged so we can say,
$\frac{F 2}{F 1}$ $\propto$ $\left({\left(v 2\right)}^{2} / {\left(v 1\right)}^{2}\right)$ so $\frac{F 2}{F 1}$ = $\frac{36}{27}$ hence change in force $\frac{4}{3}$ rd of F1
Initial Centripetal force was $\frac{6 \cdot 88}{1}$ N i.e 54 N ( velocity calculated using 0.5 m ${v}^{2}$ = 27)
Now it has become 4/3 rd i.e 72 so change occurred by (72-54)=18 N

Jan 13, 2018

The change in centripetal force is $= 18 N$

#### 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 mass is $m = 6 k g$

The radius of the track is $r = 1 m$

The variation of centripetal force is

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

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

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

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

$= \frac{2}{1} \cdot \left(36 - 27\right) N$

$= 18 N$