Finding out applied force in order to get acceleration

[DPXJube]

Homework Statement

A person pushes a box across a horizontal floor with an initial velocity of 5.2m/s. The box has a mass of 22kg, and the coefficient of Kinetic friction between the box and the floor is 0.44. How far does the box slide before coming to rest?

Homework Equations

μk = FK/FN
Fnetx = FA - FK(assuming that right is positive and left is negative)
V2 - V1 = 2aΔd

The Attempt at a Solution

Rearrange μk = FK/FN to FK = μkFN
Find Normal Force
FN = FG
FG = mg
Get 215.6 for FG which is in turn the normal force.
FK = (0.44)(215.6)
FK = 94.9
Find acceleration
Fnet = FA - FK
ma = FA - FK

And at this point I do not know what to do as I do not have the apllied force.
The plan after that was to rearrange the Kinematic equation into Δd = V2-V1/2a and fill in for that.
Unless of course my Free Body Diagram is wrong and there IS no applied force, although that seems unlikely.

 

[ActionFrank]

Based on my reading of this problem it seems that the correct answer is the one that you hit on at the end, that there is no applied force in this problem.

 

[rock.freak667]

Homework Statement

A person pushes a box across a horizontal floor with an initial velocity of 5.2m/s. The box has a mass of 22kg, and the coefficient of Kinetic friction between the box and the floor is 0.44. How far does the box slide before coming to rest?

Homework Equations

μk = FK/FN
Fnetx = FA - FK(assuming that right is positive and left is negative)
V2 - V1 = 2aΔd

The Attempt at a Solution

Rearrange μk = FK/FN to FK = μkFN
Find Normal Force
FN = FG
FG = mg
Get 215.6 for FG which is in turn the normal force.
FK = (0.44)(215.6)
FK = 94.9
Find acceleration
Fnet = FA - FK
ma = FA - FK

And at this point I do not know what to do as I do not have the apllied force.
The plan after that was to rearrange the Kinematic equation into Δd = V2-V1/2a and fill in for that.
Unless of course my Free Body Diagram is wrong and there IS no applied force, although that seems unlikely.

Hi DPXJube,

How about we try conservation of energy since your usual method is not working

The change in kinetic energy = work done by the frictional force.

 

[DPXJube]

Based on my reading of this problem it seems that the correct answer is the one that you hit on at the end, that there is no applied force in this problem.

Unfortunately I've checked the answer for the question and there is indeed an applied force.

 

[rock.freak667]

Unfortunately I've checked the answer for the question and there is indeed an applied force.

In that case, your question cannot be solved. What was the answer given?