A model train, with a mass of #3# #kg#, is moving on a circular track with a radius of #3# #m#. If the train's kinetic energy changes from #18# #J# to #0# #J#, by how much will the centripetal force applied by the tracks change by?

1 Answer
Feb 4, 2016

The centripetal force is given by #F=(mv^2)/r#. We are given the mass and the radius and can calculate the velocity from the kinetic energy. The change is #-12# #N#.

Explanation:

The centripetal force on an object of mass #m# #kg# in circular motion at velocity #v# #ms^-1# in a circle of radius #r# #m# is given by:

#F=(mv^2)/r#

In this case we know the mass is #3# #kg# and the radius is #3# #m#, but the velocity is not immediately obvious. We are given the kinetic energy at two moments in time, though, and we know that kinetic energy is:

#E_k=1/2mv^2#

We can rearrange this to make #v# the subject:

#v=sqrt((2E_k)/m)#

Since the kinetic energy at the second time is #0# #J#, the velocity is also #0# #ms^-1#, and therefore the centripetal force is also #0# #N#.

At the first time, the kinetic energy is #18# #J#, so:

#v=sqrt((2*18)/3)=sqrt(36/3)=sqrt12# #ms^-1#

This means the centripetal force is:

#F=(mv^2)/r = (3*12)/3 = 12# #N# (since #(sqrt12)^2=12#)

So the centripetal force was #12# #N# at the beginning and #0# #N# at the end. Its change was therefore #-12# #N#.