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Nonte & Fonte are running a race in a 120 m 120 m track. Nonte's mass is 60 kg & Fonte's is 70 kg. Nonte finishes the race at 12 s 12s and Fonte finishes at 15 s 15s .Both of them had uniform acceleration.The ratio of the kinetic energies of Nonte & Fonte is x : 1 x:1 .What is x x ?


The answer is 1.34.

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1 solution

Tom Engelsman
Feb 20, 2021

For this race, Nonte & Fonte both:

1) Begin from rest ( v 0 = 0 v_{0}=0 ),

2) Travel a distance of 120 120 meters,

3) Have uniform accelerations.

Because of #3 above, we require the kinematic equations d = v 0 t + 1 2 a t 2 , v 2 = v 0 2 + 2 a d d = v_{0}t + \frac{1}{2}at^2, v^2 = v^{2}_{0} + 2ad . If we susbstitute the first expression into the second one, we obtain the expression for the final velocity: v 2 = 2 d ( 2 d t 2 ) = 4 d 2 t 2 . v^2 = 2d(\frac{2d}{t^2}) = \frac{4d^2}{t^2}.

The ratio of Nonte and Fonte's kinetic energies can be computed according to:

K E N K E F = ( 1 / 2 ) m N v N 2 ( 1 / 2 ) m F v F 2 = m N ( 4 d 2 / t N 2 ) m F ( 4 d 2 / t F 2 ) = m N t F 2 m F t N 2 = ( 60 ) ( 1 5 2 ) ( 70 ) ( 1 2 2 ) 1.339 : 1 \frac{KE_{N}}{KE_{F}} = \frac{(1/2)m_{N}v^{2}_{N}}{(1/2)m_{F}v^{2}_{F}} = \frac{m_{N}(4d^2 / t^{2}_{N})}{m_{F}(4d^2 / t^{2}_{F})} = \frac{m_{N}t^{2}_{F}}{m_{F}t^{2}_{N}} = \frac{(60)(15^2)}{(70)(12^2)} \approx 1.339 : 1

or x = 1.339 . \boxed{x = 1.339}.

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