Calculating Energy in a Box with a Flashlight

  • Thread starter Thread starter andrewm
  • Start date Start date
  • Tags Tags
    Box Energy
Click For Summary
To calculate the energy in the box from a flashlight, one can simply multiply the power of the flashlight (in Watts) by the duration it is turned on, as energy is defined as power multiplied by time. Since the scenario assumes no absorption or reflection, all emitted light energy is considered to be transmitted out of the box. The integration of power density over the box's volume is unnecessary if the total power and time are known. The key takeaway is that energy in the box can be directly calculated without needing to delve into power density calculations. This approach simplifies the problem significantly.
andrewm
Messages
49
Reaction score
0

Homework Statement



A flashlight is shining into a box. There are no reflections. We know the power of the flashlight in Watts. We know the volume of the box.

Question: How much energy (associated with the EM field) is in the box?

Homework Equations



The Attempt at a Solution



Integrate the power density (J s^-1 m^-3) over the volume of the box. But what is the power density?

Hints would be preferred over full solutions :)
 
Physics news on Phys.org
You don't need the power "density" if you already know the power! A Watt is one Joule per second. Assuming all the light energy of the is absorbed by the box, which is implied in your question, just multiply the given wattage of theflashlight by the length of time the flashlight is shined on the box.
 
The flashlight is left on indefinitely, and the light is transmitted (100%) out of the box. I should have been clear that there is no absorption or reflection.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 37 ·
2
Replies
37
Views
6K
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
903
Replies
5
Views
2K
  • · Replies 62 ·
3
Replies
62
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K