A model train, with a mass of #16 kg#, is moving on a circular track with a radius of #9 m#. If the train's kinetic energy changes from #45 j# to #9 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #16 kg#, is moving on a circular track with a radius of #9 m#.  If the train's kinetic energy changes from #45 j# to #9 j#, by how much will the centripetal force applied by the tracks change by?

Lily Adams · Accepted Answer

The change in centripetal force is #=8N#

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=16kg#

The radius is #r=9m#

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#

#=2/9*(45-9)N#

#=8N#

Benjamin Asiedu · Answer

undefined The change in kinetic energy is ΔK = K_final - K_initial = 9 J - 45 J = -36 J.
The work done by the centripetal force is W = ΔK = F * d, where d is the distance traveled along the circular path.
Since the force is always perpendicular to the direction of motion, the displacement and force are in the same direction.
So, W = F * d = F * (2 * π * r), where r is the radius of the circular track.
Solving for F, we get F = ΔK / (2 * π * r).
Substituting the given values, we get F = (-36 J) / (2 * π * 9 m) ≈ -0.637 N.
Therefore, the centripetal force applied by the tracks changes by approximately 0.637 N.