A model train, with a mass of #9 kg#, is moving on a circular track with a radius of #15 m#. If the train's kinetic energy changes from #96 j# to #48 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #9 kg#, is moving on a circular track with a radius of #15 m#.  If the train's kinetic energy changes from #96 j# to #48 j#, by how much will the centripetal force applied by the tracks change by?

Julia Aguilar · Accepted Answer

The change in centripetal force is #=6.4N#

The mass of the train is #m=9kg#

The radius of the track is #r=15m#

The centripetal force is

#F=(mv^2)/r#

#v^2=(Fr)/m#

The kinetic energy is

#KE=1/2mv^2#

Therefore,

The variation of kinetic energy is

#DeltaKE=1/2m(v_1^2-v_2^2)#

#v_1^2=(F_1r)/m#

#v_2^2=(F_2r)/m#

So,

#DeltaKE=1/2m((F_1r)/m-(F_2r)/m)=r/2(F_2-F_1)=r/2DeltaF#

#DeltaF=2/rDeltaKE#

The change in centripetal force is

#DeltaF=2/15*(96-48)=6.4N#

Carson Alexander · Answer

undefined To find the change in centripetal force applied by the tracks, you can use the formula for centripetal force: $ F_c = \frac{{mv^2}}{r} $, where $ m $ is the mass of the object, $ v $ is the velocity, and $ r $ is the radius of the circular track. Since kinetic energy is given by $ KE = \frac{1}{2}mv^2 $, you can use the given kinetic energy values to find the corresponding velocities. Then, you can calculate the centripetal force using these velocities and the given radius. Finally, find the difference between the initial and final centripetal forces to determine the change.

First, solve for the initial velocity ($ v_1 $) and final velocity ($ v_2 $) using the given kinetic energy values:

Initial kinetic energy: $ KE_1 = 96 $ J
$ KE_1 = \frac{1}{2}mv_1^2 $
$ v_1^2 = \frac{2 \times KE_1}{m} $
$ v_1^2 = \frac{2 \times 96}{9} $
$ v_1^2 = \frac{192}{9} $
$ v_1^2 \approx 21.33 $
$ v_1 \approx \sqrt{21.33} $
$ v_1 \approx 4.62 $ m/s

Final kinetic energy: $ KE_2 = 48 $ J
$ KE_2 = \frac{1}{2}mv_2^2 $
$ v_2^2 = \frac{2 \times KE_2}{m} $
$ v_2^2 = \frac{2 \times 48}{9} $
$ v_2^2 = \frac{96}{9} $
$ v_2^2 \approx 10.67 $
$ v_2 \approx \sqrt{10.67} $
$ v_2 \approx 3.27 $ m/s

Now, calculate the initial and final centripetal forces:

Initial centripetal force ($ F_{c1} $):
$ F_{c1} = \frac{mv_1^2}{r} $
$ F_{c1} = \frac{9 \times (4.62)^2}{15} $
$ F_{c1} \approx \frac{9 \times 21.33}{15} $
$ F_{c1} \approx \frac{192.03}{15} $
$ F_{c1} \approx 12.80 $ N

Final centripetal force ($ F_{c2} $):
$ F_{c2} = \frac{mv_2^2}{r} $
$ F_{c2} = \frac{9 \times (3.27)^2}{15} $
$ F_{c2} \approx \frac{9 \times 10.67}{15} $
$ F_{c2} \approx \frac{96.03}{15} $
$ F_{c2} \approx 6.40 $ N

Finally, find the change in centripetal force:
$ \Delta F_c = F_{c2} - F_{c1} $
$ \Delta F_c = 6.40 \, \text{N} - 12.80 \, \text{N} $
$ \Delta F_c = -6.40 \, \text{N} $

So, the change in centripetal force applied by the tracks is -6.40 N.