A constant force is applied, such that its vector representation is of the following formula:
In which the components are expressed in Newtons. It is used to move an object that was at point A until it reaches a point B. These points have coordinates determined in meters given by:
Determine the work done by this force on the two vector representations:
The exercise asks to determine the work, the work is composed of energy. So we have the formula that work equals the force times the distance.
The force was given, the distance too. But if you do not understand distance by points can get a bit complicated. Imagine that I am in point 5, but I left point 1, what is the distance? Simple 1 minus 5, which is 4. But if I put as Cartesian points (5,0,0) minus (1,0,0), then I will have covered (4,0,0), that is, I traveled 4 on the x axis .
Now we can see that the path is the difference between the two points, the final point B (0,1,2) and the initial A (1,0,0), then (0-1,1-0, - 2-0) which is equal to (-1,1, -2), ie walked -1 on the x-axis, 1 on the y-axis, and -2 on the z-axis.
Remember that we have the force, but let's transform the force into notation, so we have (10, 5, 2), see that we only put the values ??of the versors at the corresponding points of x and y. Now, we can do force times the distance, in multiplication by climbing. Then we have strength (10, 5, 2) and distance (-1, 1, -2). In scalar multiplication we multiply only the points of the same axes, that is, x with x, y with y, z with z.
We have 10 times -1, 5 times 1, and 2 times -2, then we add everything and we have -9, that is, the work is -9 joules.
Remember that the negative is not that you spend 9 joules, but a notation is rotated clockwise or counterclockwise.
