# A model train with a mass of 4 kg is moving along a track at 6 (cm)/s. If the curvature of the track changes from a radius of 36 cm to 45 cm, by how much must the centripetal force applied by the tracks change?

Mar 8, 2017

The change in centripeta force is $= 0.008 N$

#### Explanation:

The centripetal force is

$F = \frac{m {v}^{2}}{r}$

$m = m a s s = 4 k g$

$v = 0.06 m {s}^{-} 1$

The variation in centripetal force is

$\Delta F = {F}_{2} - {F}_{1}$

${F}_{1} = 4 \cdot {0.06}^{2} / 0.36 = 0.04 N$

${F}_{2} = 4 \cdot {0.06}^{2} / 0.45 = 0.032 N$

$\Delta F = 0.04 - 0.032 = 0.008 N$