A model train, with a mass of #2 kg#, is moving on a circular track with a radius of #3 m#. If the train's kinetic energy changes from #3 j# to #5 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #2 kg#, is moving on a circular track with a radius of #3 m#.  If the train's kinetic energy changes from #3 j# to #5 j#, by how much will the centripetal force applied by the tracks change by?

Elena Albarran · Accepted Answer

#=1.dot3N# For Circular Motion, Centripetal Force #F_c= - mr omega^2# Where #m# is the mass of the body in circular motion. #r# is the radius of the circle and #omega# is angular velocity. #-# sign means that the force is opposite to the radius vector and is directed towards the center. Now velocity #v=romega# Therefore, magnitude of the force #|F_c|=(mv^2)/r# Kinetic energy is given by #KE=1/2mv^2# Let the velocity of the train change from #v_i# to #v_f# Given is change in kinetic energy #Delta KE=1/2mv_f^2-1/2mv_i^2# #1/2mv_f^2-1/2mv_i^2=5-3=2# or #m/2(v_f^2-v_i^2)=2# or #m(v_f^2-v_i^2)=4# ........(1) Change in the centripetal force applied by the tracks #=(mv_f^2)/r-(mv_i^2)/r# #=m/r(v_f^2-v_i^2)# ........(2) Inserting value of #m(v_f^2-v_i^2)# from (1) and given value of #r# Change in the centripetal force applied by the tracks #=1/3*4# #=1.dot3N#

Parker Abbott · Answer

undefined The centripetal force applied by the tracks will change by 2 N.