Work Done on Stone: 0 Kinetic Energy Change

  • Thread starter Thread starter songoku
  • Start date Start date
  • Tags Tags
    Work Work done
AI Thread Summary
The work done on a stone with a mass of 2.5 kg, falling at a constant speed of 2 m/s in a liquid, is zero due to no change in kinetic energy. The stone reaches terminal velocity, indicating that the net force acting on it is zero. Therefore, according to the work-energy principle, the work done is calculated as zero. The discussion also includes an apology for posting in the wrong section. Overall, the conclusion is that no work is done on the stone while it moves at constant speed.
songoku
Messages
2,467
Reaction score
382

Homework Statement


A stone has mass 2.5 kg is placed in a liquid. It falls with constant speed 2 ms-1 and travels 80 cm. Find the work done


Homework Equations


W = delta Ek


The Attempt at a Solution


The work done = 0 because there is no change in kinetic energy. Am I right?

Thanks
 
Physics news on Phys.org
Oh my god..I'm so sorry I've posted oin the wrong section. Please move this thread to the appropriate section. I'm really sorry

Thanks
 
The stone is moving in liquid with terminal velocity. So the net force acting on it zero. So the work done on the stone is zero.
 
Thanks a lot rl.bhat ! :smile:

And I'm sorry for posting in wrong section.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Work done by a force going around a circle ?
Replies
3
Views
2K
B Work Done By Conservative Forces
Replies
1
Views
1K
I Is Mechanical Energy Conservation Free of Ambiguity - follow up
2
Replies
77
Views
4K
B Is the Definition of Work Done Applicable to Free Fall?
Replies
27
Views
3K
I Confused about work done on charge
Replies
8
Views
1K
B Work done in lifting and lowering an object
Replies
34
Views
4K
Work done by power supply to maintain a constant current
Replies
0
Views
540
B Conceptually understanding change in potential energy with 0 net work
Replies
9
Views
2K
Work Done in slowly pulling a thread
Replies
28
Views
1K
Work done during a collision -- Change in Kinetic Energy & change in Momentum
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top