Physics 11 – By using the LOL diagrams, students do not need to conjure up the Work-Energy Theorem. Instead, students can use first principles: by analyzing the LOL diagram and coming up with the conservation of energy equation, students can solve for any unknown. The work-energy theorem, while concise, is one extra level of abstraction that probably isn’t needed at this point. What is more important: that a student can plug numbers into the right formula or figure out what is happening with energy in the system? These aren’t mutually exclusive, but abstractions can lead to answers with understanding. This is seen all the time in math, physics, programming, etc.
For problem solving, each step is simple and achievable but these are multi-step questions. They can be difficult for students while they learn to assimilate several ideas at the same time.
Physics 11 – Working on Work was the topic for the day. We had discussed in class how working is the idea of transferring energy into a system. As well, I had introduced the idea that energy is the area under a Force-displacement graph. From this, it was relatively easy for the students to grasp that W=Fd.
The formula itself is easy to grapple with. However, the nuances need clarification. So went through 4 examples to illustrate 4 main points:
- Work is calculated from applied forces, not a net force
- Something has to move in order for there to be work. I briefly explained that even a transfer of heat is a type of work, where the “thing” that moves are all of the particles (as per the Kinetic Molecular Theory).
- We only need to consider the component of the force that is in the same direction as the movement
- Work, while being a scalar, can be represented as a negative quantity which refers to a transfer of energy out of a system.
After going through the examples, the class then did some Peer Instruction with Plicker cards. One class went from 50% correct to 98% correct. The second class improved but still need an intervention – it went from 40% correct to about 60% correct.
I won’t use this question again, though. There is in fact a small bit of horizontal force acting on the mass equal to the force of friction between the mass and the hand. If μ = 0.1, then Ffr = 0.2 N, and the work done would be 0.4 J. Only a few students asked about this, so I guess for one more year I avoided confusing too many people.