Answer:
(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;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J[/tex]
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
[tex]K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J[/tex]
the change in the kinetic energy is calculated as;
[tex]\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4[/tex]
Thus, the kinetic energy increased by a factor of 4.
