# Homework Help: Mechanical energy

1. Oct 13, 2007

### map7s

1. The problem statement, all variables and given/known data

NOTE: Use the local value of g = 9.809 m/s2
You and your roommate are dragging an exercise machine (mass 56.9 kg) down a 71.2 meter long hall, from the stair landing, where (because of union rules) the UPS driver left it, to your dorm room. The coefficient of friction with the floor is μ = 0.866. The mechanical work you do in the process is:

2. Relevant equations

ME=GPE+KE
GPE=mgy
KE=1/2 mv^2
F(friction)=mu*N
N=mg
F=ma
W=Fd

3. The attempt at a solution

I tried solving for the gravitational potential energy and then solving for the frictional force and multiplying the result by the distance...looking back on it, I'm fairly sure that it was probably the wrong method to go about this problem. I know that the mechanical energy is the sum of the potential energy and kinetic energy. However, if that is the case, I would have to solve for velocity and I am not sure how to do that with the information given. When I try to use the 2-d equations, I end up with two variables (v and vo). It would really help if I could get some help (and reasoning) about solving this problem.

2. Oct 13, 2007

### Astronuc

Staff Emeritus
Since one is moving the mass along the hallway, there is no change in elevation, therefore no change in GPE.

Apply W = Fd.

The force due to friction is simply $\mu_k$N = $\mu_k$mg.

3. Oct 13, 2007

### map7s

okay...that makes more sense....so in the equation W=Fd, for F, would I use ma where a=g? and then take that F value and subtract the force of friction from it and then multiply it by the distance?

4. Oct 13, 2007

### Astronuc

Staff Emeritus
Assume the exercise machine is traveling at constant speed, i.e. no acceleration.

The friction force is proportional to the weight (N) of the machine. N is the normal force due to gravity, which is just m*g.

The work performed is the simply the product of the friction force times the distance travel. W = Ffriction * d. Find the friction force. Please refer to my previous post.

5. Oct 13, 2007

### map7s

oh....okay...that actually makes sense now....thank you so much! that was really helpful!