A model train, with a mass of #3 kg#, is moving on a circular track with a radius of #1 m#. If the train's kinetic energy changes from #21 j# to #36 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #3 kg#, is moving on a circular track with a radius of #1 m#.  If the train's kinetic energy changes from #21 j# to #36 j#, by how much will the centripetal force applied by the tracks change by?

Layla Ahmed · Accepted Answer

To make it simple, lets find out the relation of kinetic energy and centripetal force with the things we know: We know: #"K.E." = 1/2momega^2r^2# and #"centripetal force"=momega^2r# Hence, #"K.E" = 1/2xx"centripetal force"xxr# Note, #r# remain constant in the course of the process.  Hence, #Delta"centripetal force" = (2Delta"K.E.")/r = (2(36-21)J)/(1m)=30N#

Nathan Ashcroft · Answer

undefined The change in kinetic energy is equal to the work done by the centripetal force. Therefore, $ \Delta K = W = F \times d $, where $ \Delta K $ is the change in kinetic energy, $ W $ is the work done, $ F $ is the force, and $ d $ is the distance traveled.

The work done by the centripetal force is given by $ W = F \times d = F \times 2\pi r $, where $ r $ is the radius of the circular track.

So, we have $ \Delta K = F \times 2\pi r $.

We know that the change in kinetic energy $ \Delta K $ is $ 36 \, \text{J} - 21 \, \text{J} = 15 \, \text{J} $.

Substituting the values, we get $ 15 = F \times 2\pi \times 1 $, which simplifies to $ F = \frac{15}{2\pi} $ N.

Therefore, the centripetal force applied by the tracks changes by $ \frac{15}{2\pi} $ N.