Mechanical Energy of 56.9kg Exercise Machine in 71.2m Hall

AI Thread Summary
The discussion revolves around calculating the mechanical work done while dragging a 56.9 kg exercise machine down a 71.2-meter hallway, considering a coefficient of friction of 0.866. Participants clarify that gravitational potential energy is not a factor due to no change in elevation. The key formula for work is established as W = F_friction * d, where the friction force is determined by F_friction = μ * N, with N being the normal force (N = mg). The conversation emphasizes that the exercise machine moves at a constant speed, implying no acceleration. Ultimately, understanding the relationship between friction, normal force, and work is crucial for solving the problem accurately.
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Homework Statement



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:

Homework Equations



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

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.
 
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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_kN = \mu_kmg.
 
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?
 
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.
 
oh...okay...that actually makes sense now...thank you so much! that was really helpful!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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