A model train, with a mass of #9 kg#, is moving on a circular track with a radius of #15 m#. If the train's kinetic energy changes from #72 j# to #48 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 #15 m#.  If the train's kinetic energy changes from #72 j# to #48 j#, by how much will the centripetal force applied by the tracks change by?

Emily Abbate · Accepted Answer

The change in centripetal frce is #=3.2N#

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 is #r=15m#

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#

#=2/15*(72-48)N#

#=3.2N#

Charles Aeschliman · Answer

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

Fc = (mv^2) / r

Given:
Initial kinetic energy (Ki) = 72 J
Final kinetic energy (Kf) = 48 J
Mass (m) = 9 kg
Radius (r) = 15 m

We know that kinetic energy is directly proportional to the square of velocity (v). So, we can use the ratio of kinetic energies to find the ratio of velocities:

(Kf / Ki) = (vf^2 / vi^2)

Now, we can rearrange this equation to solve for the final velocity (vf):

vf = vi * sqrt(Kf / Ki)

Given that kinetic energy is given by the formula K = (1/2) * m * v^2, we can find the initial and final velocities:

vi = sqrt((2 * Ki) / m)
vf = sqrt((2 * Kf) / m)

Now, we can substitute the values and find the initial and final velocities:

vi = sqrt((2 * 72) / 9) ≈ 8 m/s
vf = sqrt((2 * 48) / 9) ≈ 6 m/s

Now, we can find the initial and final centripetal forces:

Fci = (m * vi^2) / r
Fcf = (m * vf^2) / r

Substituting the values, we get:

Fci ≈ (9 * 8^2) / 15 ≈ 38.4 N
Fcf ≈ (9 * 6^2) / 15 ≈ 21.6 N

Now, we can find the change in centripetal force:

ΔF = Fcf - Fci
ΔF ≈ 21.6 - 38.4
ΔF ≈ -16.8 N

Therefore, the centripetal force applied by the tracks will decrease by approximately 16.8 N.