Work done on moving a submerged object

Click For Summary
SUMMARY

The discussion focuses on calculating the work done on a submerged object using principles of buoyancy and tension. The buoyant force (Fb) was calculated as 9800 N using a density (rho) of 1000 kg/m³, gravitational acceleration (g) of 9.8 m/s², and a volume (V) of 1 m³. The gravitational force (Fg) of the tank was determined to be 49 N. The net force equation (Fnet = Fb - Fg - T = 0) was used to find the tension (T) as 9751 N, leading to the calculation of work done as 9751 N multiplied by 50.4 m. However, the importance of including units in calculations was emphasized, indicating potential deductions in scoring for missing units.

PREREQUISITES
  • Understanding of buoyancy principles
  • Knowledge of force calculations (Fg and Fb)
  • Familiarity with net force equations
  • Basic grasp of work-energy principles
NEXT STEPS
  • Study the concept of buoyant force in fluid mechanics
  • Learn about tension in static and dynamic systems
  • Explore work-energy theorem applications in physics
  • Review the importance of units in scientific calculations
USEFUL FOR

Students in physics, educators teaching fluid mechanics, and anyone involved in solving problems related to forces and work done in submerged systems.

Murray Inglis
Messages
2
Reaction score
0
Homework Statement
A 5.00 kg air-filled, sealed, rigid float tank that has a volume of 1.00 m3
is pulled 50.4 m
down to the seafloor in order to assist in lifting a sunken object. A diver standing on the
seafloor cranks a winch to pull the tank down.
How much work is required to pull the tank down? (You may safely ignore the mass
in the air tank and the mass of the winch rope.)
Relevant Equations
Fb = p(rho)Vg
I first worked out the buoyant force using rho = 1000, g = 9.8 and V = 1. (this gives Fb = 9800)
I then worked out Fg of the tank (5 x 9.8 = 49)
Then I used Fnet = Fb - Fg - T = 0 to work out tension as 9751.
I thought the work done would be the work done by tension (9751 x 50.4), but I'm not sure if this works.
 

Attachments

  • Screenshot 2021-09-27 122851.png
    Screenshot 2021-09-27 122851.png
    6 KB · Views: 180
Physics news on Phys.org
Murray Inglis said:
I thought the work done would be the work done by tension (9751 x 50.4), but I'm not sure if this works.
Sounds good to me.
 
Doc Al said:
Sounds good to me.
Oh great, thanks. It was a 7 mark question so it just felt like I was missing something.
 
Murray Inglis said:
Oh great, thanks. It was a 7 mark question so it just felt like I was missing something.
You are missing the units next to the numbers you provided. If I were scoring this, I would subtract at least 1 mark for this.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
29
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
37
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
3K