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?

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Answer 1 · 2017-07-11T14:15:40

The change in centripetal force is #=1.07N#

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 · 2017-07-11T14:35:51

#DeltaF_c~~-1.0667 N#

#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