Introduction to Work, do I have the right concept?

[DavidAp]

I'm just learning work in my physics class and, for some reason, I'm having difficulty understanding it. The problem in my homework assignment asks:

A particle moves 5 m in the positive x direction while being acted upon by a constant force F = <4,2,-4>.The work done on the particle by this force is:

I know that there are several useful equations which I should use when it comes to work,
W = ∫F*dr
K = 1/2mv^2
Ug = mgy
Fs = -kx
Us = 1/2kx^2

but I don't know how to apply any of these equations to this problem. I tried the first but was unsure of what F should be because F consists of i+j+k. I found the magnitude of F and plugged it into F but then didn't know what to do with the integral... so I got ride of it and just multiplied it to 5m.

I got the wrong answer so I tried with another approach. Again, I got ride of the integral because it's indefinite and I wasn't sure on how to find the constant (I am taking this integral to literally? Is it more symbolic than mathematical?) but this time I decided to make F = 4 because, in the vector <x, y, z> 4 was the x, or i, component of the equation, and I got the right answer!

So my question is, does this apply to all cases? If I was given dr and the vector force <x,y,z> does the work depend the direction the particle moves through?

For example,
F = <1,2,3>, dr = 5m.

Particle moves in the x-direction, W = 5J.
Particle moves in the y-direction, W = 10J.
Particle moves in the z-direction, W = 15J.

Thank you for taking the times to review my question.

 

[PhanthomJay]

In your problem and related examples, which are essentially correct, the Force is constant, hence, when you take the integral, it boils down to F.r (the DOT product of 2 vectors F and r, the definition of Work.). The DOT product of the two vectors is FRcos theta, where theta is the angle between the 2 vectors. You might be able to see that when a force is perpendicular to the displacement , it does no work, because theta is 90 degrees and thus cos theta is 0. Only the force in the direction of the displacement does work.

 

[cepheid]

You don't need any of the equations you listed except for the first one. The rest are mostly irrelevant, and apply to gravity, springs, or other things that we don't have in this problem. Believe it or not, you need only the first equation, which is the definition of work. But be careful. It should actually be the following:

$$W = \int_{\textrm{path}} \vec{F} \cdot d \vec{r}$$

So what you're integrating is the dot product of the force vector with the infinitesimal displacement vector. Notice that this is a special type of higher-dimensional integral known as a "line integral" or a "path integral" in which the integration proceeds in 3D space along the path traveled by the particle. Do not worry though. In your case, the line integral will reduce to a regular 1D integral. The infinitesimal displacement vector is basically the vector that corresponds to moving a really small distance along the direction of the particle path and can be expressed like this:

$$d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$$

However, in this case, the displacement is entirely in the x-direction, so the displacement vector reduces to:

$$d\vec{r} = dx \hat{i}$$

I think you should have no problem taking the dot product of the force vector with this. The path is also 1D here, and just corresponds to a range in x:

$$W = \int_0^5 \vec{F} \cdot \hat{i} dx$$

 

[DavidAp]

You don't need any of the equations you listed except for the first one. The rest are mostly irrelevant, and apply to gravity, springs, or other things that we don't have in this problem. Believe it or not, you need only the first equation, which is the definition of work. But be careful. It should actually be the following:

$$W = \int_{\textrm{path}} \vec{F} \cdot d \vec{r}$$

So what you're integrating is the dot product of the force vector with the infinitesimal displacement vector. Notice that this is a special type of higher-dimensional integral known as a "line integral" or a "path integral" in which the integration proceeds in 3D space along the path traveled by the particle. Do not worry though. In your case, the line integral will reduce to a regular 1D integral. The infinitesimal displacement vector is basically the vector that corresponds to moving a really small distance along the direction of the particle path and can be expressed like this:

$$d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$$

However, in this case, the displacement is entirely in the x-direction, so the displacement vector reduces to:

$$d\vec{r} = dx \hat{i}$$

I think you should have no problem taking the dot product of the force vector with this. The path is also 1D here, and just corresponds to a range in x:

$$W = \int_0^5 \vec{F} \cdot \hat{i} dx$$

Thank you guys so much for your help and thorough explanations! I really appreciate it!

 

[asteriatic]

Dear student,

First of all, it's great that you are actively thinking and trying different approaches to understand the concept of work. It is a fundamental concept in physics and it takes time and practice to fully grasp it. Let me try to explain it to you in a way that may help you understand it better.

Work is defined as the force applied on an object multiplied by the distance the object moves in the direction of the force. In other words, work is a measure of the energy transferred to or from an object by a force. This means that the direction of the force and the direction of the object's motion should be the same for work to be done.

In the problem you mentioned, the force F = <4,2,-4> is acting on the particle in the positive x direction. This means that the particle is moving in the same direction as the force, so work is being done on the particle. The work done by this force can be calculated using the equation W = F * d, where d is the distance the object moves. Since the particle moves 5m in the positive x direction, we can plug in the values and calculate the work done as follows:

W = (4i + 2j - 4k) * (5i)
= (20i + 10j - 20k)

The result is a vector quantity, which means that the work done has both magnitude and direction. In this case, the direction of the work done is the same as the direction of the force and the object's motion, which is the positive x direction. Therefore, we can write the final answer as W = 20 J.

To answer your question, yes, the direction of the object's motion does affect the work done. In the example you gave, if the particle is moving in the x-direction, the work done will be 5 J. But if the particle moves in the y-direction, the work done will be different (10 J) because the direction of the force and the direction of motion are not the same. This is why it is important to pay attention to the direction of the force and the object's motion when calculating work.

In terms of the equations you listed, they are all useful in different scenarios. The equation W = ∫F*dr is used when the force is not constant, meaning it changes as the object moves. The other equations you mentioned, such as K = 1/2