Does the normal force do work?

[endaman95]

Homework Statement

Consider the mass m1 on a simple incline plane. It passes from point A downwards with the velocity v0 and stops at point B (still on the plane).

What is the work done by the normal force on mass m during the complete motion?

Homework Equations

The Attempt at a Solution

In my book it says zero but I don't think it is because the mass moves diagonally down which shows movement in x and y axis. Since the normal force acting on the mass isn't perpendicular to the y-axis movement. It should do work in the opposite direction. What do you think?

 

[Doc Al]

In my book it says zero but I don't think it is because the mass moves diagonally down which shows movement in x and y axis. Since the normal force acting on the mass isn't perpendicular to the y-axis movement. It should do work in the opposite direction. What do you think?

Work it out. What are the vertical and horizontal components of the normal force and the displacement? Figure out the work done by each component and add together to get the total.

Your book is correct. Note that the normal force is perpendicular to the displacement down the incline. But you'll get the same answer when you break it into vertical and horizontal components.

 

[HallsofIvy]

You seem to be under the impression that there is a "work in the y-direction" and "work in the x-direction". There isn't. Work is a scalar quantity, not a vector.

 

[endaman95]

Thank you. My misconception was thinking of it as a vectoral quantity. Now I get it.

 

[Doc Al]

You seem to be under the impression that there is a "work in the y-direction" and "work in the x-direction". There isn't. Work is a scalar quantity, not a vector.

You can certainly choose to look at the components of a force separately and consider the work done by each to get the total. (Of course you must add them up!)

 

[paisiello2]

Work is a scalar quantity, not a vector.

While it may not be a vector there can still be a direction associated with work in the same way that kinetic energy is also a scalar quantity but the velocity still has a direction associated with it. Direction in this sense then doesn't necessarily mean a vector but rather that it is a function of the x and y variables.

 

[haruspex]

While it may not be a vector there can still be a direction associated with work in the same way that kinetic energy is also a scalar quantity but the velocity still has a direction associated with it. Direction in this sense then doesn't necessarily mean a vector but rather that it is a function of the x and y variables.

To put some algebra around that notion, let the (normal) force vector be expressed as ##\vec F = F_x\vec i + F_y \vec j## and the displacement vector by ##\vec r = r_x\vec i + r_y \vec j##. Then the work done is ##\int \vec F.d\vec r = \int F_x.dx + \int F_y.dy##.
In the present problem, ##\int F_x.dx ## and ## \int F_y.dy## are equal and opposite, so cancel.