A model train, with a mass of $5 k g$ $5 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's kinetic energy changes from $25 j$ $25 j$ to $40 j$ $40 j$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-04-15T05:49:32

The change in the centripetal force is $= 15 N$ $=15N$

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

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

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

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

$=(2)/r*Delta KE$

$=2/1*(40-25)N$

$=15N$

A model train, with a mass of 5 kg5kg5 kg, is moving on a circular track with a radius of 1 m1m1 m. If the train's kinetic energy changes from 25 j25j25 j to 40 j40j40 j, by how much will the centripetal force applied by the tracks change by?