A model train, with a mass of $6 k g$ $6 kg$ , is moving on a circular track with a radius of $4 m$ $4 m$ . If the train's kinetic energy changes from $27 j$ $27 j$ to $42 j$ $42 j$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2017-10-13T05:24:33

The change in centripetal force is $= 7.5 N$ $=7.5N$

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 mass is $m=6kg$

The radius of the track is $r=4m$

The variation of centripetal force is

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

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

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

$=(2)/r*Delta KE$

$=2/4*(42-27)N$

$=7.5N$

A model train, with a mass of 6 kg6kg6 kg, is moving on a circular track with a radius of 4 m4m4 m. If the train's kinetic energy changes from 27 j27j27 j to 42 j42j42 j, by how much will the centripetal force applied by the tracks change by?