Share
Usain Bolt’s world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bolt accelerat
Question
Usain Bolt’s world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bolt accelerated from rest at a rate of 9.50 m/s^2 for the first 0.890 s, and eventually reached a top speed of 12.4 m/s by exerting an average horizontal force of 820 N against the ground for the entire 9.58 s duration of the race. a. What was the average horizontal force (in N) exerted by Bolt against the ground during the first 0.890 s of the race?b. What was Bolt’s speed (in m/s) after the initial acceleration phase?c. What was the power expended by Bolt during the initial acceleration phase?d. It has been shown that, because of his large frame and 6’5″ height, Bolt experienced significant drag forces during the sprint. To estimate the energy lost to drag forces during the race, let’s focus on the remaining 9.58 s − 0.890 s = 8.69 s of the race after Bolt’s initial burst of acceleration. The drag force varies with speed, but let’s find an average drag force over the last 8.69 s of the race. Model the drag force as a friction force and find the increase in internal energy of Bolt and the surrounding air in these 8.69 s as Bolt runs through the air.e. Finally, find the power that Bolt must expend just to overcome the drag force and compare it to the result in part (c).
in progress
0
Physics
5 years
2021-08-01T19:40:57+00:00
2021-08-01T19:40:57+00:00 1 Answers
126 views
0
Answers ( )
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)
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)
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
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
where D is the air drag.
The net force can also be rewritten as
where
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
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.