A model train, with a mass of #15 kg#, is moving on a circular track with a radius of #6 m#. If the train's kinetic energy changes from #81 j# to #27 j#, by how much will the centripetal force applied by the tracks change by?

1 Answer
May 27, 2016

Answer:

The centripetal force has decreased by #18N#.

Explanation:

This question is about circular motion and how much centripetal force is being applied, which is given by the equation:

#f_c = (mv^2)/r#

where #m# is the mass of the object, #r# is the radius of the circle (the track in this case) and #v# is the tangential velocity. We have been given the mass and the radius, but not the velocity. However, we have been given the kinetic energy which is given by:

#K.E. = 1/2 mv^2#

Now, instead of solving for the velocity (and needing to take a square root, just to have to square it again, let's solve for the term #mv^2# since that appears in both equations and then substitute the result back into the first equation:

#mv^2 = 2K.E.#

substituting then gives

#f_c = (2*K.E.)/r#

We are looking for the change in centripetal force which we can write as:

#Delta f_c = f_(c2)-f_(c1) = (2*K.E._2)/r-(2*K.E._1)/r#

#Delta f_c = (2*(K.E._2-K.E._1))/r#

#Delta f_c = (2*(27J-81J))/(6m)= (-54J)/(3m)= -18N#

Therefore, the centripetal force has decreased by #18N#.

Note: The question had given us more information than we actually used - the mass was redundant since the kinetic energy alone was sufficient information. Other versions of this question may not include the mass.