Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get
W = F*AB
AB = 2j
F = 3(2i + j + 2k) = 6i + 3j + 6k
F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct?
s
That may well be true. If so, is the correct answer 2?
The displacement vector is 2j.
(2i + j + 2k) dot product 2j = 0 + 2j + 0 = 2j. So the correct answer is 2 Joules, correct?
Homework Statement
Find the work done by a force of 3 Newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).
Homework Equations
Commonly known vector operations.
The Attempt at a Solution
I found the displacement vector to be 2j...
Homework Statement
theta = tan^-1 {{v^2 - [v^4-g(gx^2)]}^1/2}/gx
I know that x is 30, theta is 45, g is 9.81 m/sec^2.
I'm trying to solve for v.
Homework Equations
The Attempt at a Solution
I've gotten all the way down to
v^2 = (v^4-86612.49)^1/2 + 4.165
What do I do from here?
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