A model train, with a mass of $3 k g$ $3 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's kinetic energy changes from $27 j$ $27 j$ to $36 j$ $36 j$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-05-07T13:12:21

The centripetal force changes by $= 18 N$ $=18N$

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

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

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

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

$=(2)/r*Delta KE$

$=2/1*(36-27)N$

$=18N$

A model train, with a mass of 3 kg3kg3 kg, is moving on a circular track with a radius of 1 m1m1 m. If the train's kinetic energy changes from 27 j27j27 j to 36 j36j36 j, by how much will the centripetal force applied by the tracks change by?