Work done on the block by force

AI Thread Summary
The discussion revolves around calculating the work done on a block being pushed across a rough surface by a force of 5.4 N, with a frictional force of 1.2 N acting against it. The block's kinetic energy increases from 4.0 J to 5.6 J over a distance of 0.5 m. Participants suggest that the work done can be determined by the change in kinetic energy plus the work done against friction. There is confusion regarding the angle of the applied force, which is clarified to be horizontal, simplifying the calculations. Ultimately, the focus is on understanding how to apply the work-energy principle in this scenario.
Zmuffinz
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Homework Statement



A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P
between A and B?


Homework Equations


W= PCos (theta)D


The Attempt at a Solution


I have no idea what to do for this!
 
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I suppose Work done on the block = change in kinetic energy + work against friction
 
Welcome to PF!

Hi Zmuffinz! Welcome to PF! :smile:

(have a theta: θ :wink:)
Zmuffinz said:
W= PCos (theta)D
…
I have no idea what to do for this!

Well, what is θ in this case? :smile:
 
well you see that's why I am stuck, i wasnt given an angle, or mass of the block or anything so I've got no clue as to what to do:(
 
songoku said:
I suppose Work done on the block = change in kinetic energy + work against friction

and how do i calculate work against friction?
 
Zmuffinz said:
well you see that's why I am stuck, i wasnt given an angle, or mass of the block or anything so I've got no clue as to what to do:(
You didn't post the diagram, but I presume that the force was applied horizontally. That will tell you all you need about angles.
 
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