Frictional work

[Function]

Hello

 

A question in preparation of my med school approval exam:

 

"A small ball with a mass of 25 g falls with a constant velocity of 0.1 ms^-1 in oil. The frictional work, done by the oil if the ball has dropped 40 cm into the oil, is:

 

A. 0.100 J

B. 0.000125 J

C. -0.100 J

D. -0.000125 J

 

---

 

Here's how I'd like to solve this:

 

According to Newton's second law:

[math]\sum{\overrightarrow{F}}=m\cdot \overrightarrow{a}[/math]

Since the ball travels with a constant velocity, [math]a[/math] and thus [math]\sum{\overrightarrow{F}}[/math] must be 0.

As far as I'm aware of, there're only 2 forces influencing the ball: gravitational force [math]\overrightarrow{F_z}[/math] ("z" for "zwaartekracht", Dutch for "gravitational force") and frictional force [math]\overrightarrow{F_w}[/math] ("w" for "wrijving", Dutch for "friction"):

 

[math]\begin{array}{rccl}& \overrightarrow{F_w} & = & -\overrightarrow{F_z}\\ \Leftrightarrow & \overrightarrow{F_w}&=&-m\cdot\overrightarrow{g}\\ & \overrightarrow{W_w}&=&\overrightarrow{F_w}\cdot\Delta x \\ \Leftrightarrow&\overrightarrow{W_w}&=&-m\cdot\overrightarrow{g}\cdot\Delta x\\ \Leftrightarrow & \left|\overrightarrow{W_w}\right|&=&-m\cdot\left|\overrightarrow{g}\right|\cdot\Delta x\\ \Leftrightarrow & W_w & = & m\cdot g\cdot\Delta x\\ \Leftrightarrow & W_w&\approx & 0.025\; kg\cdot 10\; ms^{-2}\cdot 0.4\; m\\ \Leftrightarrow & W_w & \approx & 0.100\; J\end{array}[/math]

 

(Mind the approximation of [math]\left| g\right|[/math]: no calculators are allowed on the exam)

This is what I'd pick: answer A.

I'm afraid, though, it's a bit too.. logic.

 

Could someone review this please?

 

Thanks.

 

Function

Edited April 19, 2014 by Function

[timo]

You might want to consider the sign your answer should have. In other words, what would be the effect of positive or negative work being done by the oil?

 

Note that formally the statement "A = B <=> |A| = |B|" that appears in your calculation is wrong; as readily verified by A=+1 and B=-1. Probably won't matter for med school where logic tends not to be on the curriculum. But since you posted your calculation I might as well comment on that.

[Function]

Hmm... But the force, and thus also the work, of the oil would be pointed upwards, whereas the one of the ball would be downwards; considering 'up' as positive, the force, and thus also the work of the oil should be positive, as is my answer?

[studiot]

There are two ways to look at this.

 

"Calculate the frictional work done by the oil"

 

Firstly does the oil receive energy or give energy?

 

If the oil gives energy the work done by the oil is positive.

 

If the oil receives energy the work done by the oil is negative.

 

Alternatively.

 

Work equals force times distance moved against that force.

 

The direction of the force applied by the oil is positive ie upwards, but the distance moved is downwards ie negative.

 

positive times negative make negative.

[Function]

There are two ways to look at this.

 

"Calculate the frictional work done by the oil"

 

Firstly does the oil receive energy or give energy?

 

If the oil gives energy the work done by the oil is positive.

 

If the oil receives energy the work done by the oil is negative.

 

Alternatively.

 

Work equals force times distance moved against that force.

 

The direction of the force applied by the oil is positive ie upwards, but the distance moved is downwards ie negative.

 

positive times negative make negative.

 

Ah, but of course! [math]W=F\cdot\Delta x\cdot \cos{\theta}[/math]

 

In casu: [math]W=F\cdot\Delta x\cdot\cos{180°}=-F\cdot\Delta x[/math]

 

How could I forget that

Is that cosine the key issue here? That I forgot that?

Or should [math]\Delta x[/math] also be a vector, instead of just a scalar?

Edited April 19, 2014 by Function

[studiot]

Yes in the work equation delta x is a vector.

 

The work = force . distance ........... where . is the dot product

 

Thus the cosine is unecessary, since it is inherent in the dot product of two vectors.

 

If you keep the cosine, the force and the the delta x have to be considered as scalars, but then the cosine can be positive or negative depending upon its quadrant.

Edited April 19, 2014 by studiot

[Function]

Yes in the work equation delta x is a vector.

 

The work = force x distance ........... where x is the cross product

 

Thus the cosine is unecessary, since it is inherent in the cross product of two vectors.

 

If you keep the cosine the delta x has to be considered as a scalar, but then the cosine can be positive or negative depending upon its quadrant.

 

Ah yes.. So, cosine is actually more to calculate the magnitude of W, whereas vectorial product is used to define the sense of W?

[studiot]

The force and distance act along the same line in this case so the cosine is not invoked.

 

That is the angle between the line of movement and the line of action of the force is zero and the cosine is therefore unity.

 

The positive or negative sign is relative to our statement of which way the energy flows or what force does work on what body.

 

This depends upon our statement.

 

In this case the question was what work does the oil do on the ball?

 

Well it doesn't. Where would it obtain the energy?

 

What actually happens is that the ball looses some potential energy in falling.

This energy is transferred to the oil as work done by the ball on the oil.

 

Since the oil does no work, we can call the energy it receives, by way of work done on it, negative work.

 

This idea is important because it is the cause of many errors in the application of the first law of Thermodynamics.

[Function]

The force and distance act along the same line in this case so the cosine is not invoked.

 

That is the angle between the line of movement and the line of action of the force is zero and the cosine is therefore unity.

 

The positive or negative sign is relative to our statement of which way the energy flows or what force does work on what body.

 

This depends upon our statement.

 

In this case the question was what work does the oil do on the ball?

 

Well it doesn't. Where would it obtain the energy?

 

What actually happens is that the ball looses some potential energy in falling.

This energy is transferred to the oil as work done by the ball on the oil.

 

Since the oil does no work, we can call the energy it receives, by way of work done on it, negative work.

 

This idea is important because it is the cause of many errors in the application of the first law of Thermodynamics.

 

Ah, very well then, the energy that the ball loses and thus the energy the oil receives is 0.025 kg * 10 ms^-2 * (-0.4) m = -0.100 J?

 

But then the problem is that you say that "Since the oil does no work, we can call the energy it receives, by way of work done on it, negative work."

 

So the "work" done by the oil should be the opposite, thus 0.100 J?

If this is not the case, and 0.100 J is the amount of energy received by the oil, I'd say that W = E₁-E₂ = 0 J - 0.100 J = -0.100 J

 

It's getting rather complicated now, and I think I'll just stick with forces: it is still the frictional work, done by the oil which is asked... It seems best to me just to work with frictional force then, and distance

Edited April 20, 2014 by Function

[studiot]

I'm trying to avoid the thorny isse that strictly speaking the answer to the question

 

"What work is done by the oil ?"

 

is zero.

 

So strictly speaking the question is faulty since it does not offer zero in the list of options.

 

The next part is really a language question.

 

Although the oil does not do any work, there is work done. This work is done by the ball.

 

Both the ball and the oil cannot do work on each other or the two amounts of work would cancel out and there would be not net energy exchange.

 

If you wish to consider this in terms of forces then the ball rubs against the oil as it falls and in so doing applies a frictional force to the oil.

 

Also in falling the ball moves its point of application of this frictional force.

 

Remember there are two conditions for a force to do work. Both these must be met.

Not only must a force be applied but it must also move its point of application..

 

Now the oil also applies a frictional force on the ball

But the oil does not fall.

So the oil does not move the point of application of the frictional force on the ball.

So the work done by the oil = force times zero movement = zero work.

 

So the only work done is by the ball and energy passes from the ball to the oil.

 

What the question wants is for you to calculate this energy and call it negative since it is from the ball to the oil.

Edited April 20, 2014 by studiot

[Function]

Ah yes. Very well. Thank you. I will try to remmeber this for the exam.