Where do the variables fit in?

Question

In the problem:

Find the work done by a force F=-7k (magnitude 7 newtons) in moving an object along the line form the point (1,2,1) to the origin (distance in meters).

I think it's supposed to fit this equation

Work=||F|| ||AB|| Cos Q,
and I think that F=7,
but where do you get ||AB|| and the Q?

Answer 1

F = 0i + 0j - 7 k

||F|| = 7

||AB|| = V (1^2 + 2^2 + 1^2 ) = sqrt 6

Its the distance between the origine and the point (1,2,1)

In general:

||AB|| = V[(x_A - x_ B)^2 + (y_A - y_ B)^2 + (z_ A- z_B)^2]

To calculate cos Q, imagine a box whose sides are 1, 2 and 1.

The diagonal of the base is V (1^2 + 2^2) = V5

The diagonal of the box is V [(V5)^2 + 1^2] = sqrt (5 + 1) = V 6

The triangle that the diagonal of the base and the diagonal of the box and one of its sides (the opposite to the angle one) is a right one. Since the vector F is parallell to the z axis, the angle is the one that form this axis with the diagonal of the box. The adjacent side is then 1, and the hypothenuse is V6

Hence cos Q = 1/V6.

So, now lets plug these values in the Work equation

W = 7* V 6* 1/V6 = 7 Joule, since the force is in Newton and the distance is in Meter.

Ana

Answer 2

Work (by definition) is force applied through a distance, so you need to know the distance traveled and the magnitude of the force applied in the direction of the travel. In the equation you give, ||F|| is the absolute magnitude of the force, ||AB|| is the absolute magnitude of the distance travelled (from A to B) and cosQ is the cosine of the angle Q between the displacement vector (AB) and the applied force vector (F). In your problem, since you've given the force as -7k, I'm assuming that you're working in the usual (i, j, k) co-ordinate space. Since the point is located at (1,2,1) the distance (||AB||) from the point to the origin is √(1²+2²+1²) = √6. The angle Q we need is the angle between the k-axis and the straight line from (1, 2, 1) to the origin. Actually, what we need is the cosine of that angle which is (by definition) the side adjacent divided by the hypotenuse. In this case, the side adjacent is 1 and the hypotenuse is √6 so cosQ = 1/√6 and the total work is
W = (7)*(√6)*(1/√6) = 7 Joules.

HTH

Doug