Work done by a constant force problem

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Homework Help Overview

The problem involves calculating the work done by a constant force on a block being moved at a constant speed over a surface with friction. The block's mass is given, along with the coefficient of kinetic friction and the angle at which the force is applied.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the applied force and the angle used in the work formula. Questions arise regarding the determination of the normal force due to the vertical component of the applied force.

Discussion Status

Participants are actively exploring the problem, questioning assumptions about the normal force and the calculations leading to the work done. There is a recognition of the need to analyze vertical and horizontal forces separately, and some guidance has been provided on how to approach the problem.

Contextual Notes

There is a noted complexity in the problem due to the vertical component of the applied force affecting the normal force, which some participants are attempting to clarify. The original poster expresses confusion regarding the calculations and the discrepancy with the textbook answer.

mooney82
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Homework Statement


A 1.8kg block is moved at constant speed over a surface for which uk = 0.25. The displacement is 2 m. It is pushed with a force at 45 degrees below the horizontal.
Find the work done by: (a) the force F; (b) friction; (c) gravity


Homework Equations


W=Fs cos (theta)

s=displacement


The Attempt at a Solution



First I found the force F to be 6.24 N.

Then using W=Fs cos (theta) I put in:

W=6.24 N * 2 m * cos 135
W= -8.82

The answer in the back of the book says 11.8, I'm wondering where I'm making my mistake.
 
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mooney82 said:
First I found the force F to be 6.24 N.
How did you solve for the force?


Then using W=Fs cos (theta) I put in:

W=6.24 N * 2 m * cos 135
Why did you use an angle of 135? (That's why your answer is negative.) But first figure out the force properly.
 
Well, I summed the x component forces:
Fcos45 - friction = ma

acceleration = 0
friction = uk(N)
N equals mg

so

Fcos45 - uk(mg)=0

F = uk(mg)/cos45

F = 0.25*1.8*9.81/cos45
 
mooney82 said:
N equals mg
Here's the problem. In this case, N ≠ mg. The fact that the applied force has a vertical component changes the normal force.

To find the normal force, analyze the vertical components of the forces acting on the block.
 
Doc Al said:
Here's the problem. In this case, N ≠ mg. The fact that the applied force has a vertical component changes the normal force.

To find the normal force, analyze the vertical components of the forces acting on the block.

Would you then solve by substitution? I'm having a hard time because F seems to cancel out.
 
mooney82 said:
Would you then solve by substitution?
Yes. You'll have two equations: one for vertical forces; one for horizontal. You can eliminate N by substituting one into the other.
I'm having a hard time because F seems to cancel out.
Show what you're doing.
 
Doc Al said:
Yes. You'll have two equations: one for vertical forces; one for horizontal. You can eliminate N by substituting one into the other.

Show what you're doing.

I got it to work out. You tha man Al!
 

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