# A model train, with a mass of 2 kg, is moving on a circular track with a radius of 3 m. If the train's kinetic energy changes from 4 j to 15 j, by how much will the centripetal force applied by the tracks change by?

Dec 26, 2016

The centripetal force will increase by $1.25 N$ from $1.33 N$ to $2.58 N$

#### Explanation:

Since centripetal force is based on the mass of the train, the radius of the track and the speed of the train, we need to find this final quantity. We can do this using the kinetic energy:

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

$4 = \frac{1}{2} \left(2\right) {v}_{i}^{2}$

${v}_{i} = 2 \frac{m}{s}$

This means the initial centripetal force is

${F}_{c i} = \frac{m {v}_{i}}{r} = \frac{\left(2\right) \left(2\right)}{3} = \frac{4}{3} N \approx 1.33 N$

At a similar way, we find the final speed from the final kinetic energy:

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

$15 = \frac{1}{2} \left(2\right) {v}_{f}^{2}$

${v}_{f} = \sqrt{15} \frac{m}{s}$

This means the centripetal force at this time is

${F}_{c f} = \frac{m {v}_{f}}{r} = \frac{2 \left(\sqrt{15}\right)}{3} \approx 2.58 N$

The change in centripetal force will be an increase of

$2.58 - 1.33 = 1.25 N$