Finding out applied force in order to get acceleration

AI Thread Summary
The discussion revolves around calculating the distance a box slides before coming to rest when pushed across a floor. The box has an initial velocity of 5.2 m/s, a mass of 22 kg, and a coefficient of kinetic friction of 0.44. Participants analyze the forces acting on the box, including the normal force and kinetic friction, but encounter confusion regarding the applied force needed for calculations. It is suggested to use conservation of energy as an alternative approach, but there is disagreement about whether an applied force exists in the problem. Ultimately, clarification is sought on the correct answer to the problem, indicating that an applied force is indeed present.
DPXJube
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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.
 
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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.
 
DPXJube said:

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 :wink:

The change in kinetic energy = work done by the frictional force.
 
ActionFrank said:
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.
 
DPXJube said:
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?
 
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