A model train, with a mass of #9 kg#, is moving on a circular track with a radius of #4 m#. If the train's kinetic energy changes from #33 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 #9 kg#, is moving on a circular track with a radius of #4 m#.  If the train's kinetic energy changes from #33 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 #=1.5N#

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 mass is #m=9kg#

The radius of the track is #r=4m#

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#

#=2/4*(36-33)N#

#=1.5N#

Parker Abbott · Answer

undefined To find the change in centripetal force, we first need to calculate the initial and final velocities of the train using the given kinetic energies. Then, we can use the formula for centripetal force to find the initial and final forces, and finally, subtract the initial force from the final force to determine the change.

First, we find the initial and final velocities using the kinetic energy formula:

Initial kinetic energy (KE1) = 33 J
Final kinetic energy (KE2) = 36 J

KE = 1/2 * mass * velocity^2

Initial kinetic energy:
33 = 1/2 * 9 * v1^2
v1^2 = (33 * 2) / 9
v1^2 = 66 / 9
v1^2 ≈ 7.333
v1 ≈ √7.333
v1 ≈ 2.71 m/s

Final kinetic energy:
36 = 1/2 * 9 * v2^2
v2^2 = (36 * 2) / 9
v2^2 = 72 / 9
v2^2 ≈ 8
v2 ≈ √8
v2 ≈ 2.83 m/s

Now, we find the initial and final centripetal forces using the formula:

Centripetal force (F_c) = mass * velocity^2 / radius

Initial centripetal force:
F_c1 = 9 * 2.71^2 / 4
F_c1 ≈ 19.447 N

Final centripetal force:
F_c2 = 9 * 2.83^2 / 4
F_c2 ≈ 21.294 N

Change in centripetal force:
ΔF_c = F_c2 - F_c1
ΔF_c ≈ 21.294 - 19.447
ΔF_c ≈ 1.847 N

Therefore, the centripetal force applied by the tracks changes by approximately 1.847 N.