Trouble Comprehending Energy and Work

[Fletcher]

This is my first attempt to find help online and my first time at this forum, and there are a lot of sections here. I hope I located this topic correctly.

This is from my physics book. I'm not currently taking any classes, so this isn't a homework question per se.

A particle moves along an x axis. Does the kinetic energy of the particle increase, decrease, or remain the same if the particle's velocity changes (a) from -3m/s to -2m/s and (b) from -2m/s to 2m/s? (c) In each situation, is the work done on the particle positive, negative, or zero?

OK, -3m/s to -2m/s seems simple enough. Some negative work is done because the velocity decreases, lowering the kinetic energy.

My problem is with the change of direction in the second case. The book says the kinetic energy remains the same and the work done is zero. I get this, but according to W = Fd, for the W to equal 0 the displacement must be 0 or the F must be 0. Clearly the F isn't 0 since the particle changed direction. So the displacement must be 0.

Since the displacement is 0, does this mean the particle instantly changed direction, or that it went some to the left and then that same amount to the right? Or could it be either?

 

[berkeman]

It is confusing, but there are two ways to think about it. KE is not a vector, it is a scalar, so it doesn't matter which way the particle is moving. If it has the same magnitude of velocity, it has the same KE. And if the KE is the same between two situations, then no work was necessary to affect the change in KE.

Now, here's a practical way to carry out the change of direction for the particle without doing any work. As the particle moves by, attach a taut string to the particle, with the other end of the taut string firmly attached to some anchor point. Attach the string so that it is perpendicular to the velocity vector of the particle right at the moment of attachment. This makes the particle all of a sudden start uniform circular motion around the anchor point of the string. Just wait the appropriate amount of time for the velocity vector of the particle to be changed to the new desired direction, and release the particle from the string. The centriptal force from the string is always directed normally to the velocity vector of the particle (true of all uniform circular motion), so no work is done on the particle by the string. But as you can visualize, the direction of the particle's motion can be changed arbitrarily. Same scalar velocity before and after, same KE throughout, no work done on the particle. Makes sense?

PS -- Welcome to PF, Fletcher.

 

[andrevdh]

For motion along a line the sign of the displacement and velocity indicates the direction of motion of the object. For positive quantities of these it is moving to the right and for negative values to the left. In the second case the particle moved for part of the time to the left and then changed direcion and moved to the right (assuming there was a continuous change in its velocity due to a constant force to the right changing its velocity in the positive x-axis direction). That means for the first part of its motion the work was negative and for the second part it was positive.

 

[Doc Al]

My problem is with the change of direction in the second case. The book says the kinetic energy remains the same and the work done is zero. I get this, but according to W = Fd, for the W to equal 0 the displacement must be 0 or the F must be 0. Clearly the F isn't 0 since the particle changed direction. So the displacement must be 0.

Since the displacement is 0, does this mean the particle instantly changed direction, or that it went some to the left and then that same amount to the right? Or could it be either?

Assuming that the force on the particle is constant (same magnitude and direction; parallel to the particle's motion), the net displacement must be zero since the net work is zero.

Imagine that you are pushing to the right against a box sliding with some speed to the left. While the box slows down, both of you are moving to the left (during this time you're doing negative work, as andrevdh explained); when the box finally stops, then starts moving to the right, both of you move to the right. Since the box ends up going at the same speed as it started, you know that the net work you've done must be zero, and that you must have ended up at back where you started--your net displacement must be zero. But you certainly moved: first left, then right.

 

[Fletcher]

Thanks for the replies. I thought that two separate acts of work could result in a net work of 0, but the problem seemed to imply to me that there was one instance of work, which I guess there is. But there's no way/reason to represent it as such (?). (If that makes any sense.) So I began to wonder about instantaneous changes in kinetic energy or direction, which I guess doesn't exist/matter anyway.