A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at this instant? KEi = J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=

(a)  the runner's kinetic energy at the given instant is 308 J

(b)  the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

the change in the kinetic energy is calculated as;

Thus, the kinetic energy increased by a factor of 4.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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