Finding the coefficient of friction on a ramp

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Homework Help Overview

The problem involves determining the coefficient of friction between a trunk and a ramp, given the work done while pushing the trunk at a constant speed along an inclined plane. The ramp is inclined at an angle of 22 degrees, and the work done is quantified as 2.2 kJ over a distance of 3.1 m.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of force based on work and distance, questioning the method of dividing work by distance to find force. There is also a focus on the implications of constant speed and how it relates to the forces acting on the trunk.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made in the original attempt. Some guidance has been offered regarding the relationship between energy changes and work, but no consensus has been reached on how to proceed without the mass of the trunk.

Contextual Notes

There is a noted absence of the mass of the trunk in the problem statement, which is critical for solving for the coefficient of friction. Additionally, participants clarify the distinction between the distance moved and the speed of movement, which may impact the understanding of the problem.

Mastevvs
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Homework Statement



You do 2.2 kJ of work pushing a trunk at a constant speed 3.1 m along a ramp inclined upward at 22 degrees. What is the frictional coefficient between the trunk and the ramp?

Homework Equations



W = F*d
Ff = μmg * cos(θ)
Fg = mg * sin(θ)

The Attempt at a Solution



To find the force exerted on the trunk up the ramp, I divide 2200 J by 3.1 m to get 710 N.

Since the trunk is moving at a constant speed, the forces pulling the trunk down the ramp must equal the forces pushing it up the ramp.

So,

710 = Fg + Ff

710 = mg*sin(22) + μmg*cos(22)

710 = mg( sin(22) + μcos(22) )

...Aaaaand this is where I get stuck. I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing. My book gives 0.6 as the answer.

Any help would be greatly appreciated.
 
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Hey you told it moves up with constant speed 3.1m/s
then why are you dividing work with speed to get the force
 
Viru.universe said:
Hey you told it moves up with constant speed 3.1m/s
then why are you dividing work with speed to get the force
possibly:
1. that would not give force - work/speed has dimensions of force per unit time.
2. the time the work was done in was not given
3. the "3.1" figure is "3.1m" - not "3.1m/s". Spot the difference.

Mastevvs said:
I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing.
Welcome to PF;
Indeed: there is information provided in the problem that has not been used in the attempt at a solution. eg. it says that the movement was at constant speed - what does this tell you about the forces?

The way to handle this sort of problem is, usually, though conservation of energy.
Start by describing the energy changes that happen and recall how this is related to the work.
But in this case you have done enough groundwork already to get away without it.
 
Last edited:
Mastevvs said:
710 = mg( sin(22) + μcos(22) )

...Aaaaand this is where I get stuck. I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing. My book gives 0.6 as the answer.

Any help would be greatly appreciated.

Your work is correct and the problem can not be solved without the mass given.

ehild
 

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