# Work With Regards to Velocity

1. May 22, 2008

### ddelaiarro

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.

2. May 22, 2008

### gstrosx

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.