How Does Water Affect the Net Force on a Submerged Stone?

[vcsharp2003]

How? The net force is downwards and displacement is upwards.

Since the $PE$ of stone increases when it's moved up, so net work done should be postive.

 

[Steve4Physics]

Excuse me jumping-in but here are a few thoughts.

@rudransh verma, try following these rules:

Put an arrow over the symbol for a vector, e.g. $\vec V$. (But for unit vectors, use a ‘hat’, e.g. $\hat i$.)

To represent the magnitude of a vector, use the unadorned letter, e.g. $V$. Or if required for extra clarity, use $|\vec V|$.

Never write equations where one side is a vector and the other side is a scalar. It is simply wrong.

I think the above is consistent with what Resnick does in your image in Post #45.

For example, take $\hat j$ as the positive upwards (+y) unit vector. Suppose there are two forces on an object, $\vec A$ up and $\vec B$ down. Then we can write these vector equations:
$\vec F_{net} = \vec A + \vec B$
$\vec F_{net} = A \hat j + B(-\hat j)$
$\vec F_{net} = (A - B)\hat j$
Since $\vec F_{net}$ has only a y-component we can write
$F_{net}\hat j= (A - B)\hat j$

Cancelling out $\hat j$ gives this scalar equation:
$F_{net} = A - B$
________________________

Also worth noting: the original problem in Post #1 should ask for the minimum amount of work. This minimum would be achieved by raising the stone very slowly so that viscous (drag) forces and any increase in kinetic energy can be ignored.

Edit. Typo's corrected. Also, my post is a bit late as the thread has progressed.

 

[rudransh verma]

Since the $PE$ of stone increases when it's moved up, so net work done should be postive.

Well why not $\Delta U=-W$
Since W=-250J so $\Delta U= 250J$ which is +ve as it should be. Meaning U increases.

 

[vcsharp2003]

Well why not $\Delta U=-W$
Since W=-250J so $\Delta U= 250J$ which is +ve as it should be. Meaning U increases.

Change in $PE$ is simply $mgh$. What is change in $KE$? What you've in above post is wrong since the equation you gave is the relationship between $PE$ and work done by the corresponding conservative force i.e. by mg in this case. You need to use the equation I mentioned and you need to understand the meaning of net work done.

 

[Doc Al]

How? The net force is downwards and displacement is upwards.

What's lifting the stone? That's the force that's doing the work.

 

[Steve4Physics]

@rudranch verma, while the stone is being lifted, from your free body diagram, you will see that there are three forces acting (assuming negligible drag):
1) Weight, $\vec W$ downwards
2) Buoyancy $\vec B$ upwards
3) The applied external lifting force, $\vec L$ upwards (e.g. the tension from a piece of string tied to the stone).

You are being asked to find the work done by $\vec L$, not the work done by $\vec W+ \vec B$.

Assuming the stone is raised at a constant velocity (no acceleration) note that
$\vec F_{net} = \vec W$ + $\vec B$ + $\vec L$ = 0
Or in terms of magnitudes
$F_{net} = -W + B + L = 0$

 

[rudransh verma]

You are being asked to find the work done by L→, not the work done by W→+B→.

What's lifting the stone? That's the force that's doing the work.

L=mg-rhoVg (taking g=10)
W=Ld
W=250 J
If it was in air then the applied force would have done W=500J.
As it should be. In air we have to do more work.

I would use the following equation to solve this problem.

$W_{net}= \Delta KE + \Delta PE$, where $W_{net}$ is work done by applied forces (excluding force of gravity)

I don't know about this formula. All I know is $\Delta KE +\Delta U=0$ for a closed system.

 

[vcsharp2003]

All I know is ΔKE+ΔU=0 for a closed system.

This equation is valid only if no forces other than gravity force act on the object. In your question, the object has an upthrust force as well as an applied force acting as the object is brought up to the surface. Therefore, your equation cannot be applied to the scenario in question.

The equation that I gave is a general form of work energy theorem.

 

[rudransh verma]

The equation that I gave is a general form of work energy theorem.

Ok

 

[vcsharp2003]

Ok

It makes problem solving easier when dealing with problems under Work,Energy and Power chapter.