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20 November 2021

An ill-posed problem involving the work-energy theorem

A source of physics questions showed a diagram like the one to the right.  A person can exert a 150 N force parallel to the frictionless ramp.  How much work does this person do on the 20 kg box to bring it from the bottom to the top of the ramp?

Approach 1:  Define the system as the box-only.  Ignore the work done by the earth, because the problem asks only for the work done by the person on the box.  Because the force of the person is parallel to the ramp, the work done is just force times distance - (150 N)(5 m) = 750 J.

Approach 2:  Define the system as the box-earth.  The work done by the person should equal the potential energy gain of the box-earth system.  That's mgh, where h is the 2.5 m vertical displacement (i.e. 5 m times sine of 30 degrees).  The work done is thus (20 kg)(10 N/kg)(2.5 m) = 500 J.

Which is correct?  

Neither.  But the problem is ill-posed.

The easiest way to see the ill-posediness is to try to draw an energy bar chart.  Let's use the box-earth system.

The gravitational potential energy bars are easy - from zero at the bottom of the ramp to something at the top.  

On an energy bar chart, conservation of energy is written by equating the number of bars on the left plus the bars of work done by an external force with the number of bars on the right - I describe the process in this video as "bars plus bars equal bars."  Right away, you can see why the problem is ill-defined!  What's going on with the kinetic energy?

We could assume constant speed.  Then the bar chart shows that we need the same amount of work done by the person as the gravitational energy at the top of the ramp.  That's a totally reasonable assumption!  This gives the work done by the person as 500 J, as in approach 2.  

But then what of approach 1?  The 150 N force applied over 5 m does give 750 J of work done!  But that would cause the box to have 250 J of kinetic energy at the top of the ramp!  Is that okay?  Well, sure, but wouldn't the box fly off the top of the ramp, then?!  I suppose, the problem said the person "can" exert 150 N of force, so perhaps the person is applying less force than they "can"?!?  But then I'm sounding like a lawyer, so by definition I'm wrong!!!

Creating multiple exclamation points in a blog post means the problem is ill posed.  Now, "ill-posed" doesn't mean "One student personally can't figure out the answer."  Were such a problem to accidentally show up on an in-class or standardized exam, I'd expect a student to explain the issue in writing - not to come to the front of the room to argue with the proctor. 

Most frequently, this sort of question asks the minimum amount of work the person needs to do to bring the box to the top of the ramp without specifying the force applied by the person.  Then the assumption of constant speed is not just reasonable but required to find the minimum work.  It's the additional specification of the force of the person on the box that causes the issue.

That doesn't mean this scenario couldn't be used for a well-posed problem!   "Explain why this problem is unsolvable without knowing how the box changes speed on the ramp" would be an excellent question.  Or, remove the "frictionless" statement, specify that the block is at rest at the bottom and the top, and ask for a justification of whether the ramp is frictionless or not.


19 November 2021

Describing motion from position-time graphs

 

The position-time graph to the right represents the motion of a cart on a track.  The motion detector points to the left.  Explain how to reproduce this graph.

When describing motion, students only need to address two elements: 

a. Which way is the cart moving?
b. Is the cart speeding up or slowing down?

Yet, students trying to answer this question will tie themselves in knots addressing how the cart is "accelerating", trying to use words like displacement, velocity, vector, negative, positive, etc.

Don't even give your class the string with which to tie knots.  Get them in the habit of addressing each of these two questions only; and get them starting every response with a fact of physics.

For the initial foray into position-time graphs, only three facts are relevant: 
  1. A position-time slope like a front slash / means the object is moving away from the detector.
  2. A position-time slope like a back slash \ means the object is moving toward the detector.
  3. To determine how fast an object is moving, look at the steepness of the position-time graph.

a. Which way is the cart moving?  Do NOT accept an answer that begins with "The cart is moving right."  I don't care what else this response says, it is wrong on its face.  A response must begin with a fact of physics.*

*"But that's not fair, Greg! A student who says "the cart is moving west because the slope is negative" is right!"  See, I'm not concerned with whether this particular answer is right or wrong - I'm concerned with my students developing a long-term deep understanding of position-time graphs, such that they can handle any question on a high-stakes exam as easily as Serena Williams handles a shoulder-high volley at the net.  Imaging Serena as a wee lass asking someone to hit her volleys for practice... and that someone kept lobbing her.  "Please hit me volleys."  "But I won the point!" says her suddenly FORMER practice partner.

"A position-time slope like a back slash \ means the object is moving toward the detector.  This graph is always a back slash, so the cart moves opposite the way the detector is pointing - the cart moves RIGHT."

b. Is the cart speeding up or slowing down?  Similarly, do not accept any answer that doesn't start with a fact of physics - especially do not accept an answer that references acceleration.  Yes, it is technically possible for an experienced physicist with a deep understanding of mathematical physics to recognize that the concave-down graph means the second time derivative is negative, and the negative slope means negative velocity, and to connect that to the magnitude of the velocity vector getting larger.  Aarrgh!  No!  No introductory physics student thinks this way!*

* And the vanishingly rare unicorn who can in fact think this way should have no trouble whatsoever using the much simpler facts above to reason through this graph.  So make the unicorn do so.

"To determine how fast an object is moving, look at the steepness of the position-time graph.  This graph is always getting steeper, so the cart always speeds up."

c. Now go reproduce the graph with a motion detector and cart on a track.  Amazingly, even after answering these two questions correctly, about 20% of the class will still set up the situation incorrectly.  They'll claim they need a curved track.  They'll have the cart moving away from the detector because "the graph is sloped down", even though they just wrote clearly on their own paper that the cart moves toward the detector!

Let them mess up.  

When the students have trouble reproducing this graph, they're confronting their personal misconceptions.  Even if a friend just shows them what to do, they'll see for themselves, "oh, the cart had to move toward the detector and speed up!  Now I get it!"  Or, you can ask them to read back to you what they wrote: "Which way is the cart moving, again?  And is the cart speeding up or slowing down?  So did you set up the cart moving toward the detector and speeding up?"

I don't like having students use their hands or their bodies to reproduce motion.  You'll end up in arguments about whether the graph does or doesn't look like it's supposed to, because it's really tough to keep a hand or a whole body continually speeding up for a second or two.  No, use a cart on a track!  This can be done with a fan cart on a flat track, or with a free-wheeling cart on a slanted track.  Insist on seeing one full second of motion - then you won't see just the 0.1 s when the student pushed the cart to get it started on the incorrect motion, and you won't have to argue about why that's wrong.  It's pretty much impossible to do this wrong and get a good-looking graph that's at least one second long.