A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #2 m#. If the train's kinetic energy changes from #72 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 #8 kg#, is moving on a circular track with a radius of #2 m#.  If the train's kinetic energy changes from #72 j# to #36 j#, by how much will the centripetal force applied by the tracks change by?

Elena Albarran · Accepted Answer

The change in centripetal force is #=36N#

The centripetal force is

#F=(mv^2)/r#

The kinetic energy is

#KE=1/2mv^2#

The variation of kinetic energy is

#Delta KE=1/2mv^2-1/2m u^2#

#=1/2m(v^2-u^2)#

The variation of centripetal force is

#DeltaF=m/r(v^2-u^2)#

#DeltaF=2m/r1/2(v^2-u^2)#

#=(2)/r*1/2m(v^2-u^2)#

#=(2)/r*Delta KE#

#=2/2*(72-36)N#

#=36N#

Sebastian Acosta · Answer

undefined To find the change in centripetal force, we'll first calculate the initial and final speeds of the train using the given kinetic energies. Then, we'll use the formula for centripetal force to determine the initial and final forces, and finally, find the difference.

The formula for kinetic energy is $ KE = \frac{1}{2} mv^2 $.
Given that the initial kinetic energy ($ KE_1 $) is 72 J and the mass ($ m $) is 8 kg, we can find the initial speed ($ v_1 $).

$ 72 = \frac{1}{2} \times 8 \times (v_1)^2 $
$ v_1 = \sqrt{\frac{2 \times 72}{8}} $
$ v_1 = \sqrt{18} $
$ v_1 = 4.24 \, m/s $

Similarly, for the final kinetic energy ($ KE_2 $) of 36 J, we can find the final speed ($ v_2 $).

$ 36 = \frac{1}{2} \times 8 \times (v_2)^2 $
$ v_2 = \sqrt{\frac{2 \times 36}{8}} $
$ v_2 = \sqrt{9} $
$ v_2 = 3 \, m/s $

Now, we can find the initial and final centripetal forces ($ F_1 $ and $ F_2 $) using the formula $ F = \frac{mv^2}{r} $.

For the initial force ($ F_1 $):
$ F_1 = \frac{8 \times (4.24)^2}{2} $
$ F_1 = \frac{8 \times 18}{2} $
$ F_1 = \frac{144}{2} $
$ F_1 = 72 \, N $

For the final force ($ F_2 $):
$ F_2 = \frac{8 \times (3)^2}{2} $
$ F_2 = \frac{8 \times 9}{2} $
$ F_2 = \frac{72}{2} $
$ F_2 = 36 \, N $

The change in centripetal force ($ \Delta F $) is the difference between the initial and final forces:
$ \Delta F = F_2 - F_1 $
$ \Delta F = 36 \, N - 72 \, N $
$ \Delta F = -36 \, N $

Therefore, the centripetal force applied by the tracks will change by -36 N.