Question

A model train, with a mass of `4 kg`, is moving on a circular track with a radius of `15 m`. If the train's kinetic energy changes from `32 j` to `24 j`, by how much will the centripetal force applied by the tracks change by?

Answer 1

The change in centripetal force is `=1.07N`

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 radius is `=15m`

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`

`=2/15*(32-24)N`

`=1.07N`

Answer 2

`DeltaF_c~~-1.0667 N`

Explanation:

`m=4kg`
`r=15m`
`KE_i=32J`
`KE_f=24J`

`DeltaKE=KE_f-KE_i=24-32=-8J`

`KE=1/2mv^2`

`DeltaKE=1/2mDeltav^2`

`Deltav=sqrt(2(DeltaKE)/m)`
`Deltav=sqrt((2*-8)/4)`
`Deltav=sqrt(-4)`

`F_c=ma_c`
where `F_c` is centripetal force; `a_c` is centripetal acceleration

`a_c=(v^2)/r`
Thus,
`F_c=m(v^2)/r`
`DeltaF_c=m((Deltav)^2)/r`
`DeltaF_c=4(sqrt(-4))^2/15`
`DeltaF_c=-16/15 N`
`DeltaF_c~~-1.0667 N`

Note:
Negative sign signifies the centripetal force decreases