A model train, with a mass of 8 kg, is moving on a circular track with a radius of 2 m. If the train's kinetic energy changes from 8 j to 48 j, by how much will the centripetal force applied by the tracks change by?

Jun 9, 2016

I found that it changes of $40 N$

Explanation:

I can use the definition of Kinetic Energy: $K = \frac{1}{2} m {v}^{2}$
${K}_{i} = 8 = \frac{1}{2} \cdot 8 {v}_{i}^{2}$
${v}_{i} = \sqrt{2} \frac{m}{s}$
${K}_{f} = 48 = \frac{1}{2} \cdot 8 {v}_{f}^{2}$
${v}_{f} = \sqrt{12} \frac{m}{s}$

We know that centripetal force is: ${F}_{c} = m {v}^{2} / r$
so:
${F}_{c i} = m {v}_{i}^{2} / r = 8 \frac{2}{2} = 8 N$
and:
${F}_{c f} = m {v}_{f}^{2} / r = 8 \frac{12}{2} = 48 N$

$\Delta {F}_{c} = 48 - 8 = 40 N$