A model train, with a mass of $4 k g$ $4 kg$ , is moving on a circular track with a radius of $6 m$ $6 m$ . If the train's kinetic energy changes from $12 j$ $12 j$ to $48 j$ $48 j$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-03-31T13:22:23

The centripetal force will change by $= 12 N$ $=12N$

The centripetal force is

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

The kinetic energy is

$K E = \frac{1}{2} m v^{2}$ $KE=1/2mv^2$

The variation of kinetic energy is

$Delta KE=1/2mv^2-1/2m u^2$

$=1/2m(v^2-u^2)$

The variation of centripetal force is

$DeltaF=m/r(v^2-u^2)$

$DeltaF=2m/r1/2(v^2-u^2)$

$=(2)/r*1/2m(v^2-u^2)$

$=(2)/r*Delta KE$

$=2/6*(48-12)N$

$=12N$

A model train, with a mass of 4 kg4kg4 kg, is moving on a circular track with a radius of 6 m6m6 m. If the train's kinetic energy changes from 12 j12j12 j to 48 j48j48 j, by how much will the centripetal force applied by the tracks change by?