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?

2 Answers
Jan 13, 2018

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

Jan 13, 2018

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