Work/Force question, can anyone explain?

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SUMMARY

The discussion focuses on calculating the work done on a mobile object by a force vector in a specified time interval. The force vector is defined as F = (2.80 N)i + (6.80 N)j + (7.05 N)k, and the object moves from an initial position di = (3.05 m)i - (2.00 m)j + (5.00 m)k to a final position df = -(5.00 m)i + (4.05 m)j + (7.00 m)k over 4.00 seconds. The correct method to find work involves using the dot product of the force vector and the displacement vector, which is calculated as df - di. Additionally, average power is determined by dividing the work done by the time interval.

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1. A force F = (2.80 N)i + (6.80 N)j + (7.05 N)k acts on a 2.00 kg mobile object that moves from an initial position of di = (3.05 m)i - (2.00 m)j + (5.00 m)k to a final position of df = -(5.00 m)i + (4.05 m)j + (7.00 m)k in 4.00 s.



Homework Equations



W = Fd


The Attempt at a Solution



I thought I could do the pythagorean theorem for the Force...

square root (2.8^2 + 6.8^2 + 7.05^2), that would give the magnitude of the force.

Then I did the same thing, the magnitude of df - magnitude of di.

Then, (magnitude F) * (magnitude df - magnitude di) = W

But that is incorrect. I am using Web assign online and the answer I get from the previous method does not work. Can anyone explain?
 
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Oh, I forgot the actual question haha.

(a) Find the work done on the object by the force in the 4.00 interval

(b) Find the average power due to the force during that interval.

(c) Find the angle between vectors. di and df.
 
Here is a definition of work
http://hyperphysics.phy-astr.gsu.edu/hbase/wint.html#wi

One could find it with the magnitude of force applied over the distance.

But then one simply needs to apply the dot or scalar product

F [itex]\cdot[/itex] s, where s is the difference between the position vectors df-di.

Average power is just average work (equivalent to energy) divided by the time over which the work is applied.


magnitude of df - magnitude of di
does not work in finding the distance.

The distance over which the force is applied is the magnitude of the vector which is found by taking the difference between the position vectors df - di.
 

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