a) 893 N
b) 8.5 m/s
c) 3816 W
d) 69780 J
e) 8030 W
Explanation:
a)
The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

where
m is Bolt's mass
a is the acceleration
In the first 0.890 s of motion, we have
m = 94.0 kg (Bolt's mass)
(acceleration)
So, the net force is

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).
b)
Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
In the first phase of Bolt's race we have:
u = 0 m/s (he starts from rest)
(acceleration)
t = 0.890 s (duration of the first phase)
Solving for v,

c)
First of all, we can calculate the work done by Bolt to accelerate to a speed of
v = 8.5 m/s
According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

where
m = 94.0 kg is Bolt's mass
v = 8.5 m/s is Bolt's final speed after the first phase
is the initial kinetic energy
So the work done is

The power expended is given by

where
t = 0.890 s is the time elapsed
Substituting,

d)
First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.
In the first 0.890 s, the force exerted was

We know that the average force for the whole race is

Which can be rewritten as

And solving for
, we find the average force exerted by Bolt on the ground during the second phase:

The net force exerted by Bolt during the second phase can be written as
(1)
where D is the air drag.
The net force can also be rewritten as

where
is the acceleration in the second phase, with
u = 8.5 m/s is the initial speed
v = 12.4 m/s is the final speed
t = 8.69 t is the time elapsed
Substituting,

So we can now find the average drag force from (1):

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

where
d is Bolt's displacement in the second part, which can be found by using suvat equation:

And so,

e)
The power that Bolt must expend just to voercome the drag force is given by

where
is the increase in internal energy due to the air drag
t is the time elapsed
Here we have:

t = 8.69 s is the time elapsed
Substituting,

And we see that it is about twice larger than the power calculated in part c.