Net Work Done: Solve Force Question Homework

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The discussion revolves around calculating the net work done on an object subjected to a force and friction. The force exerted is F = -4.3i + 2.0j + 1.5k Newtons, and the displacement is d = -0.8i + 1.8j - 3.6k meters. The initial calculation of work done by the force yielded 1.64 J, while the work done by friction was incorrectly calculated as 0, leading to confusion. It was clarified that the friction force of -0.2 N can be used in the work equation, resulting in an additional work of -0.82 J. The net work done is then calculated by combining both work values, but discrepancies in the final answer prompted requests for further assistance.
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Homework Statement


A force F=-4.3i+2.0j+1.5k Newtons exerts a displacement of d=-0.8i+1.8j+-3.6k metres upon a small (ie mass is negligible) object. If the friction force is equal to Ffr=-0.2,, then what is the net work done upon the object?


Homework Equations





The Attempt at a Solution


I know the net work is the sum of all the individual work and the work equation is W=F.D. So i tried to work out the work of the initial force stated by multiplying the force and displacement stated which i got 1.64J. I think the only other work is caused by the friction force but when i try to work that out (because the mass is negligable) i get 0. Can someone please help?
 
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Mass is negligible only affects calculating the friction force using \mu N, but you already have the friction force. What can you do with it?
 
could i put it as the force in the work equation and put the distance as the displacement that's stated in the question?
 
Yes.
 
so i have 1.64 from the first force i worked out and if i put the friction force into the work equation w= -0.2 x D i get -0.82. If I add the two works i have i get 0.82, but the answer is 1.3... can you give me hints on what I've done wrong??
 
I get that also. From where did you get this problem?
 

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