Does Running or Walking the Same Distance Use More Energy?

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SUMMARY

The discussion centers on whether running or walking the same distance expends more energy. The primary argument presented is based on the physics of work, defined as W = F * d, where force (F) and distance (d) are constant for both activities if mass remains unchanged. The analysis incorporates Newton's Second Law and the concept of friction, concluding that the work done in moving a mass over a distance is equivalent regardless of the speed of movement. However, additional energy expenditure factors, such as heat output, must be considered for a comprehensive understanding of total energy consumption.

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  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of work-energy principles
  • Familiarity with concepts of friction and force
  • Awareness of thermodynamics related to energy expenditure
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  • Explore the relationship between speed and energy expenditure in physical activities
  • Research the impact of body mass on caloric burn during exercise
  • Investigate the thermodynamic principles of energy transformation in human physiology
  • Learn about the kinematic coefficient of friction and its implications in various activities
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Fitness enthusiasts, exercise physiologists, sports scientists, and anyone interested in understanding the energy dynamics of physical activities like running and walking.

ddelaiarro
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You'll have to excuse me if this issue has been discussed here before. I did about 30 minutes worth of searching and didn't come upon it, so I decided to post.

A colleague and I had a lengthy discussion today on whether running a set distance spends more calories (energy) than walking that same distance.

My main argument centered around moving a mass over a distance.

My contention is that work is the product of force and distance:

W = F * d

If the same mass, m, is moved across the same distance, d, then the same force, F, is exerted. The hinge point of my argument is the assumption that your acceleration/deceleration is instantaneous.

If you look at the free body diagram, you'll note that you have force acting in two axis on the mass.

In the x-axis, you have the applied force, F, and the friction force, F_f. According to Newton's Second Law, a body in motion will stay in motion unless acted upon. Hence, in order to keep a constant velocity,

F = F_f hence \Sigma F_x=0

In the y-axis, there's a gravitational force,

F_g = m * g

and the equal and opposite normal force, F_N


F_g = F_N hence \Sigma F_y=0

The kinematic coefficient of fricition, u, is the same, regardless of velocity. The equation for friction force is:

F_f = F_N * u * cos(theta) <-- Theta being the angle at which the plane the mass is on is at. Assume a flat plain (cos0 = 1) for simplicity here.

Regardless of velocity, u and F_N are the same, hence F_f is the same.

So, if F_f is the same, F is the same. And, in the end,

W = F * d produces the same product regardless of velocity.

In the end, I convinced my coworker that my approach was correct. Although, he didn't buy the end result.

I'd love to hear some of your thoughts on this subject.
 
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Consider the heat output from someone who just ran x vs someone who just caught up to that person by walking. The work done in transporting the body is the same (assuming exactly equal body masses) however there is other energy that is spent. You have to consider total energy difference, dE = dK + d(k*Temperature) in this case.
 

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