Amok said:Are you sure that is the right answer?
marthkiki said: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.
My problem is figuring out how to represent the direction and magnitude of force as an actual vector.
The answer is (6*SQRT 5)/5 Joules. I don't understand how.
LCKurtz said:I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book.
marthkiki said: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?
The scientific definition of work is the product of force and displacement, where force is applied in the direction of the displacement. In simpler terms, work is the measure of the energy used to move an object.
To calculate work in 3 dimensions, you need to use vector components. First, find the angle between the force vector and the displacement vector. Then, use the formula W = F * d * cos(theta), where F is the magnitude of the force, d is the magnitude of the displacement, and theta is the angle between the two vectors.
The standard unit of measurement for work is joules (J). However, other units such as foot-pounds (ft-lb) and ergs (erg) are also used in different contexts.
In physics, work is directly related to energy. Work is the measure of the energy used to move an object, and energy is the ability to do work. This means that work is a form of energy.
Yes, the work done on an object can be negative. This occurs when the force and displacement are in opposite directions, resulting in a negative value for work. This indicates that the object gained energy rather than losing it.