A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #8 m#. If the train's kinetic energy changes from #4 j# to #15 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 #8 m#.  If the train's kinetic energy changes from #4 j# to #15 j#, by how much will the centripetal force applied by the tracks change by?

Isabella Alvarez · Accepted Answer

The centripetal force change by #=2.75N#

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#

#=2/8*(15-4)N#

#=11/4N#

#=2.75N#

Adam Arnold · Answer

undefined To find the change in centripetal force, we can use the formula:

ΔF = Δ(1/2 mv^2) / r

Where:
ΔF is the change in centripetal force,
Δ is the change in kinetic energy,
m is the mass of the train (4 kg),
v is the velocity of the train,
r is the radius of the circular track (8 m).

First, we find the change in velocity using the change in kinetic energy:

ΔK = 1/2 mv^2 - 1/2 mu^2

15 J - 4 J = 1/2 * 4 kg * v^2 - 1/2 * 4 kg * u^2

11 J = 1/2 * 4 kg * v^2 - 1/2 * 4 kg * 0

11 J = 1/2 * 4 kg * v^2

22 J = 4 kg * v^2

v^2 = 22 J / 4 kg

v^2 = 5.5 m^2/s^2

v ≈ √(5.5) m/s
v ≈ 2.345 m/s

Now, we calculate the initial centripetal force:

F_initial = mv^2 / r

F_initial = (4 kg * (2.345 m/s)^2) / 8 m

F_initial ≈ 2.348 N

Now, we calculate the final centripetal force:

F_final = (4 kg * v^2) / 8 m

F_final ≈ (4 kg * (2.345 m/s)^2) / 8 m

F_final ≈ 2.348 N

Finally, we find the change in centripetal force:

ΔF = F_final - F_initial

ΔF ≈ 2.348 N - 2.348 N

ΔF ≈ 0 N