# A model train with a mass of 5 kg is moving along a track at 14 (cm)/s. If the curvature of the track changes from a radius of 7 cm to 24 cm, by how much must the centripetal force applied by the tracks change?

A change of $\Delta F = \frac{119}{120} = 0.9916666 \text{ }$Newton

#### Explanation:

Given data: $v = 14 \text{ }$cm/second$= 0.14 \text{ }$meter/second
$m = 5 \text{ }$kilogram
${r}_{1} = 7 \text{ }$centimeter$= .07 \text{ }$meter
${r}_{2} = 24 \text{ }$centimeter$= 0.24 \text{ }$meter

Centripetal Force formula

${F}_{1} = \frac{m {v}^{2}}{r} _ 1 = \frac{5 \cdot {\left(.14\right)}^{2}}{.07} = \frac{7}{5} \text{ }$Newtons

${F}_{2} = \frac{m {v}^{2}}{r} _ 2 = \frac{5 \cdot {\left(.14\right)}^{2}}{.24} = \frac{49}{120} \text{ }$Newtons

${F}_{1} - {F}_{2} = \Delta F = \frac{119}{120} = 0.9916666 \text{ }$Newton

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