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

Answer:

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

Answer:

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#