Question

A sprinter who weighs 706 N runs the first 5.5 m of a race in 1.6 s, starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the 1.6 s? (c) What average power does the sprinter generate during the 1.6 s interval?

Answer 1

Answer:

(a) Speed = 6.875 m/s

(b) Kinetic Energy = 1.70 kJ

(c) Average power =  1.06 kW

Step-by-step explanation:

Given:

d = 5.5 m

t = 1.6 s

W = 706 N

Part (a)

Using the equation of motion with the assumption of zero initial speed:

$d = \frac{1}{2} ({v-0})t$

$5.5= \frac{1}{2} (1.6v)$

The speed of the sprinter then become:

$v = 6.875 m/s$

Part (b)

Convert the weight of the sprinter to mass:

$m = \frac{w}{g}$

$m = \frac{706}{9.8}$

$m = 72 kg$

KE (Kinetic Energy) can then be calculated as:

$KE = \frac{1}{2} mv^{2}$

$KE = \frac{1}{2} (72)(6.875)^{2}$

$KE = 1701.56 J = 1.70 kJ$

Part (c)

The change of the kinetic energy is equivalent to the work done by the sprinter. The average power P is the rate of the work done and can be calculated as

$P = \frac{KE}{t}$

$P = \frac{1701.56}{1.6}$

$P = 1063.48 W = 1.06 kW$