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## Main Question or Discussion Point

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:

If the same mass,

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,

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

and the equal and opposite normal force,

The kinematic coefficient of fricition,

Regardless of velocity,

So, if

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.

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 [tex]\Sigma[/tex]*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 [tex]\Sigma[/tex]*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.