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3 years ago

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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=______________

1 answer:

Liono4ka [1.6K]3 years ago

4 0

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;

$K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J$

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

$K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J$

the change in the kinetic energy is calculated as;

$\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4$

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

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