Finding work by Force over Position graph

[Jrlinton]

Homework Statement

An object of 5 kg is acted on by a force and moved from 0.5 to 2.0m.
How much work does the force do?
As the object moved, friction with a coefficient of 0.2 also acted upon it. What is its final speed if it started at rest?

Homework Equations

The Attempt at a Solution

So the work done from 0.5m to 2.0 m was easy enought to find as I just used the graph to find the area between the data and the x-axis coming up with 27.5J

I calculated the work of friction (-u*m*g*d) -0.2*5kg*9.81m/s/s*1.5m= -14.715 J. I subtracted this from the 27.5 joules found earlier to get 12.785 J. Set 12.785=0.5*m*v^2=(0.5)(5kg)(v^2) and found v to be 2.26 m/s. This was incorrect...[/B]

 

[Simon Bridge]

Some observations on what you wrote:

Your graph shows force as a function of time (from t=0 to t=2sec) - you have written "s" and "sec" for units on that axis.

If you subtract -14.715J from 27.5J, then you get 42.215J

The friction force is a constant 9.8N ... how do you know your calculation is incorrect (ie. if computer mediated, maybe you were supposed to use 10N/kg for gravity).

 

[Jrlinton]

I apologize, the units should be meters. The graph is given from 0m to 2m but the work is the be calculated from 0.5m to 2m. The work of the friction, as it opposes the work of force F, should be added, I stated that i subtracted meaning the magnitude of the fictional work.

 

[haruspex]

Any idea what the significance is of the section of the graph from 0 to .5m? Was that for another part of the question? I ask in case for the second part you are supposed to take the object as starting from rest at position 0.

 

[Simon Bridge]

Second above - looking at your graph, where did the object start?