Why do some physics textbooks say P = dK/dt and others say P = dE/dt?

  • Thread starter Thread starter member 731016
  • Start date Start date
  • Tags Tags
    Physics Textbooks
Click For Summary
The discussion centers on the definitions of power (P), kinetic energy (K), and energy (E) in physics textbooks, questioning why some texts use P = dK/dt while others use P = dE/dt. It clarifies that power represents any energy transfer, including work done by external forces, leading to the equation P = dW/dt. There is confusion regarding whether E is equated to K in modern physics, despite the understanding that E = K + U (potential energy). The conversation emphasizes the importance of context in defining energy types and their relationships. Overall, the distinctions in equations reflect different interpretations of energy transfer in physics.
member 731016
Homework Statement
I am interested in why some physics textbooks say P = dK/dt (namely morins classical mechanics) while others say that P = dE/dt (namely physics for scientists and engineers with modern physics)
Relevant Equations
P = dK/dt = dE/dt
Are they assuming that E = K in physics for scientists and engineers with modern physics, but I though E = K + U?Many thanks!
 
Last edited by a moderator:
Physics news on Phys.org
What is P, K and E?

Is P power, is K kinetic energy and E is energy, what kind of energy?
 
  • Like
Likes member 731016
malawi_glenn said:
What is P, K and E?

Is P power, is K kinetic energy and E is energy, what kind of energy?
Thanks for your reply @malawi_glenn! Here I'll send you a screen shot what it says.
 
1670133901931.png
 
The text pretty much explains it.

The generic definition of power is ANY energy transfer, like transfer of heat etc.

Then in the case of an external force that is performing work, you will get P = dW/dt
 
  • Like
Likes member 731016
malawi_glenn said:
The text pretty much explains it.

The generic definition of power is ANY energy transfer, like transfer of heat etc.

Then in the case of an external force that is performing work, you will get P = dW/dt
Ok thank you @malawi_glenn!
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...