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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 #21 j#, by how much will the centripetal force applied by the tracks change by?

Answer 1

Nora Arzate

The change in centripetal force is #=6N#

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*(21-12)N#

#=6N#

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Answer 2

Julia Aguilar

The change in centripetal force is #=6N#

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*(21-12)N#

#=6N#

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Answer 3

Charles Aeschliman

The change in kinetic energy of the model train can be calculated as ΔKE = KE_final - KE_initial.

Given that the initial kinetic energy (KE_initial) is 12 J and the final kinetic energy (KE_final) is 21 J, the change in kinetic energy (ΔKE) is 21 J - 12 J = 9 J.

The change in centripetal force (ΔF) can be calculated using the formula: ΔF = ΔKE / r, where r is the radius of the circular track.

Substituting the values, ΔF = 9 J / 3 m = 3 N.

Therefore, the centripetal force applied by the tracks changes by 3 Newtons.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.