A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #3 m#. If the train's kinetic energy changes from #12 j# to #6 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #3 m#.  If the train's kinetic energy changes from #12 j# to #6 j#, by how much will the centripetal force applied by the tracks change by?

William Audy · Accepted Answer

The change in centripetal force is #=4N#

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#

#=2/3*(12-6)N#

#=4N#

Elena Albarran · Answer

undefined The change in kinetic energy is equal to the work done by the centripetal force, so:

ΔKE = W = FΔd

The change in kinetic energy is:

ΔKE = 6 J - 12 J = -6 J

The work done is:

W = FΔd

Since the work done is negative, it means the force opposes the motion.

The distance traveled by the train along the circular track during this change in kinetic energy can be calculated using the formula for the circumference of a circle:

Δd = 2πr

Δd = 2π(3 m) = 6π m

Now, we can calculate the force:

F = ΔKE / Δd

F = (-6 J) / (6π m)

F ≈ -0.318 N

So, the centripetal force applied by the tracks changes by approximately 0.318 N.

Carson Alexander · Answer

undefined The change in kinetic energy is equal to the work done by the centripetal force. The work done by the centripetal force is equal to the change in kinetic energy. Therefore, the centripetal force decreases by $6 \, \text{J}$.