Question

A model train, with a mass of `6 kg`, is moving on a circular track with a radius of `1 m`. If the train's kinetic energy changes from `27 j` to `36 j`, by how much will the centripetal force applied by the tracks change by?

Answer 1

18 N

Explanation:

Let'assume Kinetic energy in the 1st case is K1 and in 2nd case K2
So,`(K2)/(K1)` = `(m/m) ((v2)^2 / (v1)^2)` so, ratio of square of v2 and v1 will be `36/27`
Now, Centripetal force F is given as `m(v^2)/r`
As,mass and radius are unchanged so we can say,
`(F2)/(F1)` `prop` `((v2)^2 / (v1)^2)` so `(F2)/(F1)` = `36/27` hence change in force `4/3` rd of F1
Initial Centripetal force was `(6*88)/1` N i.e 54 N ( velocity calculated using 0.5 m `v^2` = 27)
Now it has become 4/3 rd i.e 72 so change occurred by (72-54)=18 N

Answer 2

The change in centripetal force is `=18N`

Explanation:

The centripetal force is

`F=(mv^2)/r`

The kinetic energy is

`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=1m`

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`