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 $24 j$ $24 j$ to $48 j$ $48 j$ , by how much will the centripetal force applied by the tracks change by?

Question

1555 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T20:23:52

$2.05 N$ $2.05N$

The kinetic energy of the train is given by

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

which we can rearrange to find the velocity (which we need) as

$\sqrt{\frac{2 E}{m}} = v$ $sqrt((2E)/m)=v$

so, calculating the velocity in both cases ( $24 J$ $24J$ and $48 J$ $48J$ ),

$v_{1} = \sqrt{\frac{2 \times 24}{6}} = \sqrt{8} \approx 2.83 m s^{- 1}$ $v_1 = sqrt((2xx24)/6) = sqrt8 ~~ 2.83ms^-1$

$v_{2} = \sqrt{\frac{2 \cdot 48}{6}} = \sqrt{16} = 4 m s^{- 1}$ $v_2 = sqrt((2*48)/6) = sqrt16 = 4ms^-1$

Therefore,

$Deltav$ = $v_2-v_2 = 4 - 2.83 = 1.17ms^-1$

The equation of centripetal force is

$F = ma_c = (mv^2)/r$

where $F$ is force, $m$ is mass, $a_c$ is centripetal acceleration, $v$ is velocity, and $r$ is the radius of the circle.

Therefore we can say that change in centripetal force is

$DeltaF = (m(Deltav)^2)/r$

which, putting in the values we know, is

$DeltaF = (6xx1.17^2)/4 ~~ 2.05N$

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 24 j24j24 j to 48 j48j48 j, by how much will the centripetal force applied by the tracks change by?