How Much Power Does a Sprinter Generate in a Race?

Click For Summary
SUMMARY

The discussion centers on calculating the power output of a sprinter during a race, specifically a 48.0 kg sprinter running 46.0 m in 7.30 s. The horizontal force acting on the sprinter was determined to be 82.9 N using Newton's Second Law. The initial attempt to calculate power output at 1.70 s resulted in an incorrect value of 121.55 W, which was resolved by applying the formula P=F*v, yielding the correct power output. The final power output calculation reflects the relationship between force and velocity in determining a sprinter's performance.

PREREQUISITES
  • Understanding of kinematics equations, specifically x=x_0 + v_0 t + (1/2)at^2
  • Familiarity with Newton's Second Law, represented as F=ma
  • Knowledge of kinetic energy calculation using KE=(1/2)mv^2
  • Ability to compute power using the formula P=ΔE/Δt
NEXT STEPS
  • Explore advanced applications of Newton's Second Law in sports science.
  • Learn about the relationship between force, velocity, and power in athletic performance.
  • Investigate the effects of acceleration on energy expenditure in sprinting.
  • Study the biomechanics of sprinting to understand how power output varies with technique.
USEFUL FOR

Sports scientists, coaches, athletes, and anyone interested in the physics of sprinting and performance optimization.

tangibleLime
Messages
71
Reaction score
0

Homework Statement


A 48.0 kg sprinter, starting from rest, runs 46.0 m in 7.30 s at constant acceleration.

a) What is the magnitude of the horizontal force acting on the sprinter?

b) What is the sprinter's power output at 1.70 s?

Homework Equations


x=x_0 + v_0 t + \frac{1}{2}at^2
\vec{F}=ma
v=v_0 + at
KE=\frac{1}{2}mv^2
P=\frac{\Delta E}{\Delta t}

The Attempt at a Solution


For part A, I first used x=x_0 + v_0 t + \frac{1}{2}at^2 to find the acceleration of the sprinter:

x=x_0 + v_0 t + \frac{1}{2}at^2
46=0 + 0(7.3) + \frac{1}{2}a(7.3)^2
a=1.726 m/s^2

With the acceleration, I stuck it into Newton's Second Law and found the force.

\vec{F}=ma
\vec{F}=48(1.726)
\vec{F}=82.9 N

My answer for part A was correct.

Part B is where I am having some difficulties. First I got the velocity of the runner at 1.7 sec:

v=v_0 + at
v=0 + (1.726)(1.7)
v=2.934 m/s

I then calculated the amount of work done by calculating the kinetic energy, which I used because the sprinter is running and in motion.

KE=\frac{1}{2}mv^2
KE=\frac{1}{2}(48)(2.934)^2
KE=206.6 J

Then to calculate power, I took the change in the kinetic energy (0 J to 206.6 J) and divided it by the change in time.

P=\frac{\Delta E}{\Delta t}
P=\frac{206.6}{1.7}
P=121.55 W

That answer was marked incorrect. I then tried to add the horizontal force (the answer to part A), which added up to 204.45 W and that was also incorrect.

Any help would be greatly appreciated.
 
Physics news on Phys.org
Nevermind, I figured it out.

I did P=F*v which gave me the correct answer.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K