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March 2, 2015

March 2, 2015

Posted by **J** on Monday, May 7, 2012 at 11:37pm.

- Calculus -
**MathMate**, Tuesday, May 8, 2012 at 2:14pmWork done by a force F along a vector P is F.P (dot product). So total work done is

F.P + F.A

where A=<1,-3,4>, P=<7-1, 2-(-3), 5-4>

and F=15<1,2,2>/sqrt(1^2+2^2+2^2)=<5,10,10>

The division by sqrt(...) is required to normalize the direction vector to a unit vector.

- Calculus -
**J**, Wednesday, May 9, 2012 at 1:11amI don't understand it.. what is the final work done? Also, the arrows (< and >) are they actually supposed to be written as < and > or are they supposed to mean brackets ( ) or these brackets []

thank you

- Calculus -
**MathMate**, Wednesday, May 9, 2012 at 3:40pmMost books write vectors as <x,y,z> to differentiate from a point (x,y,z).

You would separate the work done by P along the two lines (vectors) OP and PA.

The force is 15N along <1,2,2>. Since <1,2,2> is not a unit vector, we need to normalize it as

<1,2,2>/sqrt(1^2+2^2+2^2)=<1,2,2>/3

So the force is

F=15<1,2,2>/3 = 5<1,2,2>=<5,10,10>

Work done is F.D, where F and D are vectors, and . represents the dot product.

The first part of the work is

F.OP=<5,10,10>.<1-0,-3-0,4-0>

=<5,10,10>.<1,-3,4>

=5*1-10*3+10*4

=15

I'll leave it to you to calculate the second part, F.PA, where

A=<7-1, 2-(-3), 5-4>=<6,5,1>

- Calculus -
**J**, Thursday, May 10, 2012 at 4:05pmcould you help me calculate the second part also? thank you

- Calculus -
**MathMate**, Thursday, May 10, 2012 at 8:53pmF.PA=<5,10,10><6,5,1>

=5*6+10*5+10*1

=30+50+1

=81

- Calculus -
**J**, Saturday, May 19, 2012 at 9:43amIsnt it 30 + 50 + 10 because 10* 1 is 10 so final is = 90

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