An object moves from the point (−2,0,2) to the point (1,15,−13). The only force acting on it is a constant F=5i−5j+5k newtons. Find the work done if the distance is measured in meters.

Answer 1

Jessica,

Work, W =F⋅d F=5i -5j+5k and d=3i+15j-15k

F⋅d=15-75-75=-135 joules

Jim

Answer 2

Hi Jessica!

This problem is best solved using a dot product between the force and displacement vectors - if you're unfamiliar with this math tool, the following explanation goes through what is happening "under the hood" when you carry out a dot product, and it gets to the same result. If you are familiar with it, feel free to skim down to the bottom**!

This problem is in three dimensions, but it can help to conceptualize it if we break it down into each component.

The work (W ) put into a system is equal to the product of the force applied (in this case - thankfully for us - a constant) and the displacement of the object in the same direction as the applied force. In one dimension, this equation looks like:

W = F * d where "d" is displacement

The problem asks for the total amount of work done in a 3D system, and we can find this by adding the work done in each component:

W_i = F_i * Δi

W_j = F_j * Δj

W_k = F_k * Δk

W_total = W_i + W_i + W_k = (F_i * Δi ) + (F_j * Δj ) + (F_k * Δk )

So in our case, our change in position looks like:

(-2, 0, 2) --> (1, 15, -13),

giving the displacement vector d

d = (Δi , Δj , Δk )

d = (1 - (-2), 15 - 0, -13 - 2)

d = (3 m, 15 m, -15 m)

We can then multiply each component of our force vector, F , with the corresponding components of the displacement vector, d, and sum the three to give the total amount of work done, W_total.

W_total = F * d = (5 N * 3 m) + (-5 N * 15 m) + (5 N * -15 m) = (15 - 75 - 75) N*m

W_total = -135 N*m = -135 J

And it's okay to have negative work done on a system!! Imagine a flying superhero is pushing against a runaway train - they're exerting a force opposite the direction of motion and the train slows down, meaning it has less mechanical energy than it had before, i.e. negative work was done to the system!

**

The process above is equivalent to taking the dot (a.k.a scalar) product between the force vector, F, and the displacement vector, d. The displacement vector is given by the changes in position in each dimension:

d = (Δi , Δj , Δk )

d = (1 - (-2), 15 - 0, -13 - 2)

d = (3 m, 15 m, -15 m)

And F was given to us as:

F = (5 N, -5 N, 5 N)

The work done to the system is then found to be:

W_total = F * d = (5 N * 3 m) + (-5 N * 15 m) + (5 N * -15 m) = (15 - 75 - 75) N*m

W_total = -135 N*m = -135 J

And it's okay to have negative work done on a system!! Imagine a flying superhero is pushing against a runaway train - they're exerting a force opposite the direction of motion and the train slows down, meaning it has less mechanical energy than it had before, i.e. negative work was done to the system!