Question

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?

Answer 1

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.