Question

Find the total work done by a 15 N force in the direction of the vector [1,2,2], when it moves a particle from O[0,0,0] to P[1,-3,4] and then from P to A[7,2,5]. Distance is measured in metres.

Answer 1

Work 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.

Answer 2

I 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

Answer 3

Most 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>

Answer 4

could you help me calculate the second part also? thank you

Answer 5

F.PA=<5,10,10><6,5,1>
=5*6+10*5+10*1
=30+50+1
=81

Answer 6

Isnt it 30 + 50 + 10 because 10* 1 is 10 so final is = 90