Work done by liquid's frictional viscosity

AI Thread Summary
A 0.75 m sphere dropped through a liquid experiences a change in energy as it falls. Initially, the sphere has potential energy, which converts to kinetic energy as it descends. The work done by the liquid's frictional viscosity opposes the sphere's motion, resulting in a lower final velocity than if it fell through air. The relationship can be expressed as initial potential energy equaling final kinetic energy plus the work done by friction. Understanding this interplay clarifies how friction affects the sphere's energy during its fall.
imatreyu
Messages
79
Reaction score
0

Homework Statement



A 0.75 sphere is dropped through a tall column of liquid. When the sphere has fallen a distance of 2.0 m, it is observed t have a velocity of 2.5 m/s. How much work was done by the frictional viscosity of the liquid?

PLEASE EXPLAIN-- I really would like to understand this concept! My review book says that intial energy is entirely potential, and it sounds like final energy is kinetic-- but isn't there PE finally as well? I'm not sure. . . Could someone just explain this situation to me? As in, what happens in terms of energy/ work/ etc?

Thank you, in advance!

Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
Potential energy is always calculated w.r.t. a reference. What your review book is doing is taking the final position of the sphere as the reference. Relative to that, the initial P.E. of the sphere is simply m g (2.0). The final P.E. is zero because it is at its reference position.

Therefore, initially energy is entirely potential, and finally it is entirely kinetic.
 
Oh thank you! I totally forgot about that.

Still though, I really don't understand what's going on in terms of the liquid's friction and work and energy in the system. What does the liquid friction work against?
 
The liquid friction works to slow down the sphere. The final velocity of the sphere is therefore lesser than it would have been had it been thrown downwards in the air. In terms of equations, this simply means

Initial P.E = Final K.E. + Work done by friction
 
Oh, thank you so much! You really cleared it up for me.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top