The discussion revolves around calculating instantaneous power from a given rate of energy increase expressed as a function of time. The problem is situated within the context of physics, specifically focusing on the relationship between work, energy, and power.
The discussion is active, with participants providing hints and questioning the assumptions made in the calculations. There is a recognition of the need to clarify the distinction between instantaneous and average power, and some guidance is offered regarding the relationship between energy and power.
Participants note the importance of understanding the definitions and units involved, particularly the relationship between joules and watts. There is an acknowledgment of the need to derive the correct expression for instantaneous power without relying on average power calculations.
Le_Anthony said:Homework Statement
The energy of a system increases at a rate of 3.5t + 6.2t^2, in joules.
What is the instantaneous power at t=3.1 s?
Homework Equations
P=dW/dt
J/s=watt
The Attempt at a Solution
dW=3.5 + 12.4t
P=(3.5 + 12.4t joule) / (3.1s)=1.12 + 4t, in watts
Yes/ no?
No, it's dW/dt = 3.5 + 12.4tdW=3.5 + 12.4t
no, they are asking for the instantaneous power at t=3.1Le_Anthony said:Homework Statement
The energy of a system increases at a rate of 3.5t + 6.2t^2, in joules.
What is the instantaneous power at t=3.1 s?
Homework Equations
P=dW/dt
J/s=watt
The Attempt at a Solution
dW=3.5 + 12.4t
P=(3.5 + 12.4t joule) / (3.1s)=1.12 + 4t, in watts
Yes/ no?
donpacino said:no, they are asking for the instantaneous power at t=3.1
why would the power at 3.1 secs be dependent on time?
1 watt does equal 1 joule per second.
I'll give you a few hints.
The average power dissipated during a time period is energy/ the time period. What happens when you decrease that time period to a very small value?
another hint. To find energy simply sum (integrate) the power. how do you go the other way?
which i now realize i was thinking of W/Δt.. oopsLe_Anthony said:So i just need to derive the equation and plug in 3.1s?
I divided by 3.1 because instantaneous power is derivative of work w/ respect to time.