# A model train, with a mass of 15 kg, is moving on a circular track with a radius of 6 m. If the train's kinetic energy changes from 81 j to 27 j, by how much will the centripetal force applied by the tracks change by?

May 27, 2016

The centripetal force has decreased by $18 N$.

#### Explanation:

This question is about circular motion and how much centripetal force is being applied, which is given by the equation:

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

where $m$ is the mass of the object, $r$ is the radius of the circle (the track in this case) and $v$ is the tangential velocity. We have been given the mass and the radius, but not the velocity. However, we have been given the kinetic energy which is given by:

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

Now, instead of solving for the velocity (and needing to take a square root, just to have to square it again, let's solve for the term $m {v}^{2}$ since that appears in both equations and then substitute the result back into the first equation:

$m {v}^{2} = 2 K . E .$

substituting then gives

${f}_{c} = \frac{2 \cdot K . E .}{r}$

We are looking for the change in centripetal force which we can write as:

$\Delta {f}_{c} = {f}_{c 2} - {f}_{c 1} = \frac{2 \cdot K . E {.}_{2}}{r} - \frac{2 \cdot K . E {.}_{1}}{r}$

$\Delta {f}_{c} = \frac{2 \cdot \left(K . E {.}_{2} - K . E {.}_{1}\right)}{r}$

$\Delta {f}_{c} = \frac{2 \cdot \left(27 J - 81 J\right)}{6 m} = \frac{- 54 J}{3 m} = - 18 N$

Therefore, the centripetal force has decreased by $18 N$.

Note: The question had given us more information than we actually used - the mass was redundant since the kinetic energy alone was sufficient information. Other versions of this question may not include the mass.