A model train with a mass of $3 k g$ $3 kg$ is moving along a track at $16 \frac{c m}{s}$ $16 (cm)/s$ . If the curvature of the track changes from a radius of $15 c m$ $15 cm$ to $18 c m$ $18 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-02-11T07:04:09

The answer is $= 2.56 N$ $=2.56N$

The centripetal force is

$F = \frac{m v^{2}}{r}$ $F=(mv^2)/r$

The variation in centripetal force is

$DeltaF=(mv^2)/(Delta r)$

$=(3*(0.16)^2)/(0.18-0.15)$

$=(3*0.16*0.16)/0.03$

$=2.56N$

A model train with a mass of 3 kg3kg3 kg is moving along a track at 16 (cm)/s16cms16 (cm)/s. If the curvature of the track changes from a radius of 15 cm15cm15 cm to 18 cm18cm18 cm, by how much must the centripetal force applied by the tracks change?