Net Work Done: Solve Force Question Homework

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SUMMARY

The net work done on an object subjected to a force of F = -4.3i + 2.0j + 1.5k Newtons and a displacement of d = -0.8i + 1.8j - 3.6k meters is calculated using the work equation W = F · d. The initial calculation of work from the force yields 1.64 Joules. The friction force, given as Ffr = -0.2 Newtons, contributes an additional work of -0.82 Joules. The total net work done is therefore 1.64 - 0.82 = 0.82 Joules, which differs from the expected answer of 1.3 Joules, indicating a potential miscalculation or misunderstanding in the application of the work equation.

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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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