Equations for fk, Wfk, and WF as an expression

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Homework Help Overview

The discussion revolves around a physics problem involving a block being pulled on a horizontal surface with friction. The original poster seeks to derive expressions for the frictional force, work done by friction, and work done by the applied force, considering the effects of an angle and the coefficient of kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express the frictional force and work done using variables like mass, distance, and angles. Some participants question the correctness of the work done formulas being used, particularly regarding the application of trigonometric functions.

Discussion Status

Participants are engaged in clarifying the expressions for the forces and work involved. There is some agreement on the expression for the frictional force, but confusion remains regarding the work done calculations. Multiple interpretations of the work done formulas are being explored, and participants are seeking to understand the discrepancies in their results.

Contextual Notes

The original poster notes that their submission to a web-based homework platform did not accept their answers, indicating potential issues with signs or calculations. There is also mention of specific values for mass, distance, angle, applied force, and the coefficient of kinetic friction, which may influence the discussion.

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Homework Statement



(a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force vector F at an angle θ with respect to the horizontal. The coefficient of kinetic friction between block and table is μk. Is the force exerted by friction equal to μkmg?
ANSWER = No (Correct )

Enter the expression that describes only the magnitude of the force exerted by friction. (Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)

fk = ?

(b) How much work is done by the friction force and by vector F ? (Don't forget the signs. Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
Wfk = ?
WF =?



The Attempt at a Solution



fk= μk (mg-Fsin(θ))
Wfk=-(cos(θ))d
WF=mg(sin(θ))d

*d= distance

I'm struggling coming up with the correct equations. Any help is appreciated.
 
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welcome to pf!

hi svfetz! welcome to pf! :smile:
svfetz said:
fk= μk (mg-Fsin(θ))
Wfk=-(cos(θ))d
WF=mg(sin(θ))d

*d= distance

your fk is correct :smile:

but i don't understand what formula you're using for work done :confused:
 
Hi :smile:

For my submission for fk webassign didn't like my answer for it so I have no idea what I am doing wrong because you agree its right.

For my other two equations I used W= FΔh and used cos and sin because it is getting pulled at an angle. So w=(Fcosθ)Δh?

Ahhh, I'm so confused right now that I don't even know what I did anymore:confused:

**For my Homework webassign hates me. (this is just the full problem)

(a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force vector F at an angle θ with respect to the horizontal. The coefficient of kinetic friction between block and table is μk. Is the force exerted by friction equal to μkmg?
No

Enter the expression that describes only the magnitude of the force exerted by friction. (Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
fk =

(b) How much work is done by the friction force and by vector F ? (Don't forget the signs. Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
Wfk =
WF =

(c) Identify all the forces that do no work on the block.
1. friction vector f k 2.vector n 3.vector F sin(θ) 4. mvector g 5.vector F cos(θ)

(d) Let m = 2.00 kg, x = 4.00 m, θ = 41.0°, F = 17.0 N, and μk = 0.400, and find the answers to parts (a) and (b). (Include appropriate signs.)

fk = -12.3 NThe response you submitted has the wrong sign.
Wfk = -4.0 J Wrong
WF = 68.15 J Wrong
 
hi svfetz! :smile:

(just got up :zzz:)
svfetz said:
For my other two equations I used W= FΔh and used cos and sin because it is getting pulled at an angle. So w=(Fcosθ)Δh?

yes :smile:
(d) Let m = 2.00 kg, x = 4.00 m, θ = 41.0°, F = 17.0 N, and μk = 0.400, and find the answers to parts (a) and (b). (Include appropriate signs.)

fk = -12.3 NThe response you submitted has the wrong sign.
Wfk = -4.0 J Wrong
WF = 68.15 J Wrong

sorry, i don't follow any of these :confused:

how did you get WF ?​
 

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