Equations for fk, Wfk, and WF as an expression

[svfetz]

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.

 

[tiny-tim]

welcome to pf!

hi svfetz! welcome to pf!

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

*d= distance

your fk is correct

but i don't understand what formula you're using for work done

 

[svfetz]

Hi

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

**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

 

[tiny-tim]

hi svfetz!

(just got up :zzz:)

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

(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

how did you get WF ?​

 

[Nickie]

For the first equation, fk, we can use the formula for friction force: fk = μkN, where N is the normal force. In this case, the normal force is equal to mgcos(θ), so the equation becomes:

fk = μk(mgcos(θ))

For the second equation, Wfk, we can use the formula for work: W = Fdcos(θ). In this case, the force is the friction force, so the equation becomes:

Wfk = fkdcos(θ) = μk(mgcos(θ))dcos(θ)

For the third equation, WF, we can use the formula for work again, but this time the force is the applied force F, so the equation becomes:

WF = Fdcos(θ) = Fdcos(θ) = (mgcos(θ))dcos(θ)

So the final expressions are:

fk = μk(mgcos(θ))

Wfk = μk(mgcos(θ))dcos(θ)

WF = (mgcos(θ))dcos(θ)