Question

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=______________

Answer 1

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

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

Step-by-step 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;

`K.E_i = (1)/(2) mu^2\\\\K.E_i = (1)/(2) * 64.1 * 3.1^2\\\\K.E_i = 308 \ J`

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

`K.E_f = (1)/(2) mu_f^2\\\\K.E_f = (1)/(2) m(2u)^2\\\\K.E_f = (1)/(2) * 64.1 \ * (2* 3.1)^2\\\\K.E_f = 1232 \ J`

the change in the kinetic energy is calculated as;

`(K.E_f)/(K.E_i) = (1232)/(308) =4`

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