Calculating Work and Speed of a Crate Pulled Across a Floor

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Homework Help Overview

The problem involves calculating the final speed of a crate being pulled across a floor with varying conditions of friction. The crate has a mass of 70 kg and is subjected to a constant horizontal force of 200 N over two segments: the first 10 m is frictionless, and the next 10 m has a coefficient of friction of 0.30.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating total work done on the crate, considering both frictionless and frictional segments. There are attempts to clarify the calculations of work and potential energy, as well as the forces acting on the crate.

Discussion Status

Some participants express uncertainty about their calculations and seek confirmation on their approach. Guidance has been offered regarding the consideration of conservative and non-conservative forces, and the potential energy associated with the system. There is an ongoing exploration of the relationship between work, energy, and the final speed of the crate.

Contextual Notes

Participants are navigating the complexities of work-energy principles and the implications of friction in their calculations. There is a focus on ensuring correct signs and values in their work calculations.

Chanster
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Homework Statement



A 70-kg, starting from rest, is pulled across a floor with a constant horizontal force of 200 N. For the first 10 m the floor is frictionless, and for the next 10 m the coefficient of friction is 0.30. What is the final speed of the crate?

Homework Equations



W= -Delta PE
Wnc= DELTA PE+DELTA KE
F=ma
Ff=MuFn
W=Fd

The Attempt at a Solution



What I tried is to calculate the total work. So I calculate the work being done frictionless.
W=FD
W=200(10)
W=2000
Now I had to calculate work on friction
W=FD
W=(686)(.30)(10)
W=2058
This is how far I have gotten.
 
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OK, so total work =?
 
I don't know if I am doing it right or not
 
Yes, you're doing it right so far.
 
total work=-58?
 
Check your signs on the two values of work you calculated, otherwise you're off to a good start. Actually, the work you calculated for the applied force is wrong. Over what distance did the 200-N force act?

Both the applied force and friction are non-conservative forces, so the work they do contribute to the lefthand side of the equation

W_{nc} = \Delta{PE}+\Delta{KE}

Now you just have to figure out the righthand side of the equation.
 
I don't understand how to calculate work at all. What kind of PE is calculated in this problem or how to calculate it
 
Oh, okay. First, determine what forces are acting on the block, and then identify which of those are conservative. Each conservative force will have a corresponding PE (though the change could very well be zero). At this point, you probably only know of two conservative forces: gravity and springs. The potential energy function for gravity is U=mgh, and for springs, it's U=1/2 kx^2.
 
vela said:
Oh, okay. First, determine what forces are acting on the block, and then identify which of those are conservative. Each conservative force will have a corresponding PE (though the change could very well be zero). At this point, you probably only know of two conservative forces: gravity and springs. The potential energy function for gravity is U=mgh, and for springs, it's U=1/2 kx^2.

Can the PE for this equation be U=MGD or is it zero since there is no height?
 
  • #10
Both! You can be pedantic and include the gravitational potential energy terms. When you plug everything in, you will find they cancel. Or you can just neglect them right from the start since you know the change in gravitational PE will be zero because the block didn't move up or down.
 
  • #11
vela said:
Both! You can be pedantic and include the gravitational potential energy terms. When you plug everything in, you will find they cancel. Or you can just neglect them right from the start since you know the change in gravitational PE will be zero because the block didn't move up or down.

Oh ok, i got 7.4 for final velocity

So I did what you told me and calculate the the force applied again i got 4k. I subtract that from 2058. I got 1942, with that I just set it equal to 1/2mv^2 and i solve for v and got 7.4?
 
  • #12
Sounds good. Don't forget the units!
 

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