What is the Force F acting on a NdYAG laser of 1 kW ?

  • Thread starter Thread starter ammar98
  • Start date Start date
  • Tags Tags
    Force Laser
Click For Summary
The discussion centers on calculating the force acting on a 1 kW NdYAG laser due to photon impulse transfer. The initial attempt uses the formula F = dp/dt, incorporating photon energy and momentum, but lacks specific values like the wavelength of the laser. Participants point out that the calculations incorrectly treat power as energy and emphasize the need for consistent notation and clarity in the equations. The correct approach requires including the laser's power in the calculations to determine the force accurately. Overall, the solution needs refinement and additional information for verification.
ammar98
Messages
1
Reaction score
0

Homework Statement


Force F acting on a NdYAG laser of 1 kW continuous wave output by the impulse transfer of the photons

Homework Equations


F= dp/dt

The Attempt at a Solution


Photon Energy E=[hc][/λ]=hν
momentum P= [h][/λ]
power of laser 1 kw
at time (t) the laser emit photons
t=[P][/E] = [λ[/hc]

F= dp/dt = t * P = [λ][/hc] * [h][/λ] = [1][/C] = 3.333 x 10 ^ -9 N

can you check please if the solution correct
 
Physics news on Phys.org
ammar98 said:

Homework Statement


Force F acting on a NdYAG laser of 1 kW continuous wave output by the impulse transfer of the photons

Homework Equations


F= dp/dt

The Attempt at a Solution


Photon Energy E=[hc][/λ]=hν
momentum P= [h][/λ]
power of laser 1 kw
at time (t) the laser emit photons
t=[P][/E] = [λ[/hc]

F= dp/dt = t * P = [λ][/hc] * [h][/λ] = [1][/C] = 3.333 x 10 ^ -9 N

can you check please if the solution correct
You need to provide the photon wavelength without which we cannot check your solution.
ammar98 said:
F= dp/dt = t * P= [λ][/hc] * [h][/λ] = [1][/C] = 3.333 x 10 ^ -9 N
If P is power (Energy per unit time) then t*P is energy not force. Also 1/c has units of Time/Length which is not units of force.
 
Where do you include the 1 kW power? It doesn't appear that you have.

Please provide numbers which you use (like wavelength of a NdYAG laser) to make it easier to check your calculations.

Another tip: try to make sure your capital P is consistent meaning Power, and momentum symbol is ρ (which you can get by clicking the Σ icon above the text box) or just use lower case p.
 
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
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
4K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K