I forgot how to do the following, so I'm not sure if they're correct or not:
Find the work done by a force "F" that causes a displacement "d".
i and j = unit vectors
a)
F = 4i + j
d = 3i + 10j
W = (4i + j)(3i + 10j)
W = 12|i|^2 + 40ij + 3ij + 10|j|^2
W = 12(1) + 40(0) + 3(0) + 10(1)
W = 22
b)
F = 2i
d = 5i + 6j
W = (2i)(5i + 6j)
W = 10|i|^2 + 12ij
W = 10(1) + 12(0)
W = 10
You want F DOT D (I am using capital for vector) = |F||D| cos T
where T is the angle between them.
dot product = scalar product = FxDx + FyDy
= 4*3 + 1*10 = 12+10 = 22
Now do that one other way
|F| = sqrt(4^2+1^2) = sqrt(17)
|D| = sqrt(3^2+10^2) = sqrt (109)
angle F to x axis = tan^-1 (1/4) =14 deg
angle D to x axis = tan^-1 (10/3)=73.3 deg
so T = 73.3 - 14 = 59.3
cosT = .5105
so
W = |F||D| cos T = 21.97 = 22 to our sig figs
check
b)
F = 2i
d = 5i + 6j
W = 2*5 + 0*6
W = 10
To find the work done by a force "F" that causes a displacement "d", you can use the formula:
W = F * d
Where:
W is the work done
F is the force vector
d is the displacement vector
Let's verify the calculations for both scenarios:
a) For the given values:
F = 4i + j
d = 3i + 10j
W = (4i + j) * (3i + 10j)
W = 12i^2 + 40ij + 3ij + 10j^2
Since i^2 = j^2 = 1,
W = 12 + 40ij + 3ij + 10
Simplifying,
W = 22 + 43ij
So, the work done for scenario a) is 22 + 43ij.
b) For the given values:
F = 2i
d = 5i + 6j
W = (2i) * (5i + 6j)
W = 10i^2 + 12ij
Since i^2 = 1,
W = 10 + 12ij
So, the work done for scenario b) is 10 + 12ij.
Please note that the work done has both a magnitude (given by the value) and a direction (given by the vector component).