Work Done query

A

apple2357

Full Member
Joined
Mar 9, 2018
Messages
546
Not sure where to post this:

I saw this question in a book:

1759430205583.png
1759430250308.png

1759430276874.png


Is this correct? Is it not the case the work done is 50J ( 20cos60 x 5) .some of that of that work (30J) goes towards overcoming friction and some of it (20J) towards increasing KE, but it's all work done. So the answer is 50J.

What does the question mean when it asks 'work done IN...' ?
 
apple2357 said:
Is this correct? Is it not the case the work done is 50J ( 20cos60 x 5) .some of that of that work (30J) goes towards overcoming friction and some of it (20J) towards increasing KE, but it's all work done. So the answer is 50J.

What does the question mean when it asks 'work done IN...' ?
Click to expand...
Yes, it is correct, the work done "IN" moving the box is 20 J.

The use of the phrase: "the work done in moving the box" means that they want you to focus only on the work that is done to move it.

"Work" is, of course, Energy and movement involves Kinetic Energy whereas resistance to movement, ie: Friction, usually results in Heat or Sound (noise); both alternative forms of Energy.

You are right that there will be (30 J of) Energy consumed (lost) in overcoming friction (as Heat and/or Sound) but the question is basically telling you to ignore that (by the very use of the phrase: "the work done in moving the box"), so you only use the resultant force (after the opposing frictional force in the direction opposite to the movement is taken into account) to calculate "the work done in moving the box".

Hope that helps. 😊
 
Last edited:
i think that helps... let me get back to you if it doesn't. I will muse.
Thanks
 
Saying just \(\displaystyle \textcolor{indigo}{\bold{Find \ the \ work \ done \ in \ moving \ the \ box.}}\) is ambiguous.

It could mean:

\(\displaystyle \bold{Work \ done \ by \ a \ specific \ force \ like \ the \ rope.}\)

Or

\(\displaystyle \bold{Net \ work \ done}\) \(\displaystyle \longrightarrow\) \(\displaystyle \bold{total \ work \ done \ by \ all \ forces.}\)


A good textbook would be more precise and would say at least one of the following:

Find the work done by the rope\(\displaystyle \cdots \longrightarrow \textcolor{blue}{\text{applied force}}\)
Find the work done against friction\(\displaystyle \cdots \longrightarrow \textcolor{red}{\text{work done by the rope}} \longrightarrow \textcolor{blue}{\text{applied force}}\)
Find the net work done\(\displaystyle \cdots \longrightarrow \textcolor{green}{\text{total work done by all forces}}\)
 
apple2357 said:
i think that helps... let me get back to you if it doesn't. I will muse.
Thanks
Click to expand...
Feel free to "muse" away but please don't get confused by anything the village idiot tells you....

logistic_guy said:
Saying just \(\displaystyle \textcolor{indigo}{\bold{Find \ the \ work \ done \ in \ moving \ the \ box.}}\) is ambiguous.
It could mean:
\(\displaystyle \bold{Work \ done \ by \ a \ specific \ force \ like \ the \ rope.}\)
Or
\(\displaystyle \bold{Net \ work \ done}\) \(\displaystyle \longrightarrow\) \(\displaystyle \bold{total \ work \ done \ by \ all \ forces.}\)
Click to expand...

This question is not in any way "ambiguous"; it is a standard Physics problem that appears in that format in every Physics text that addresses such situations.

logistic_guy said:
A good textbook would be more precise and would say at least one of the following:

Find the work done by the rope\(\displaystyle \cdots \longrightarrow \textcolor{blue}{\text{applied force}}\)
Find the work done against friction\(\displaystyle \cdots \longrightarrow \textcolor{red}{\text{work done by the rope}} \longrightarrow \textcolor{blue}{\text{applied force}}\)
Find the net work done\(\displaystyle \cdots \longrightarrow \textcolor{green}{\text{total work done by all forces}}\)
Click to expand...
That is all just rubbish!

In this case the "applied force" is 20 N which may be resolved into 10 N (being applied) in the horizontal direction and 17.32 N (to 4 s.f.) in the vertical direction.

The work done by the force in the vertical direction will (given the diagram provided) result in raising the leading edge of the box and, thereby, creating an increase in Potential (Gravitational) Energy of the system.
However, this cannot be calculated (as we are not given the Mass (or the dimensions) of the box)!

That is why you are only expected to (and can only) deal with the horizontal forces involved and the phrase: "work done IN..." tells you that you are to ignore the Energy loss due to Friction and calculate only "the work done in moving the box."

The "work done by the rope" is the "total work done" here but, for the reasons given above, this cannot be calculated.

If, after your musings, you still have any questions or you need further clarification then please do come back and ask. I'm sure that somebody (who knows what they're talking about) will address these for you.

Hope that helps. 😊
 
Last edited:
apple2357 said:
Not sure where to post this:

I saw this question in a book:

View attachment 39769
View attachment 39770

View attachment 39771


Is this correct? Is it not the case the work done is 50J ( 20cos60 x 5) .some of that of that work (30J) goes towards overcoming friction and some of it (20J) towards increasing KE, but it's all work done. So the answer is 50J.

What does the question mean when it asks 'work done IN...' ?
Click to expand...
Are you familiar with vectors and vector dot/cross products?
 
I was helping a friends son with his maths and i haven't studied too much mechanics and i realised i didn't understand it fully. I was getting confused about 'net work done' and understood work done as a scalar and not a vector ( like resultant force).
 
apple2357 said:
I was helping a friends son with his maths and i haven't studied too much mechanics and i realised i didn't understand it fully. I was getting confused about 'net work done' and understood work done as a scalar and not a vector ( like resultant force).
Click to expand...
All the Forces involved are, as you recognize, Vector quantities but the Work done (and, indeed, any Work, ie: Energy) is a Scalar quantity.

It seems rather odd that a question like this would appear in a Maths text as it definitely requires knowledge and understanding of topics that aren't normally covered in Maths courses. 🤔
 
Last edited:
Musing over....

I think if the question asked something like...

Find the work done by the tension

Or

Find the work done by each of the forces acting on the box

Or even,

Find the total work done

I would probably be less unsure...
 
apple2357 said:
Musing over....
I think if the question asked something like...
Find the work done by the tension
Or
Find the work done by each of the forces acting on the box
Or even,
Find the total work done
I would probably be less unsure...
Click to expand...
Hmmm, you don't appear to have understood what I've been trying to explain... ☹️🤷‍♂️

"the work done by the tension" cannot be calculated because the tension in the rope is 20 N but a component of that force (acting vertically) increases the Potential Energy of the system but it is not possible to calculate that increase because the Mass of the box is not given!

Neither is it possible to calculate "the work done by each of the forces acting on the box" because there is only one force acting on the box, ie: the tension in the rope (the force due to Friction is a reaction, not an applied force) and, as explained in the previous paragraph, it is not possible to calculate the total work done by the applied (20 N) force.

As I've now explained twice here and also did so previously at Post #5 (q.v.), it is simply not possible to calculate "the total work done"!

I suggest you "muse" a bit further, perhaps you might start by reading (more carefully?) all of my posts in the thread, if you really want to be "less unsure" and finally reach a full understanding of the mechanics of this question.

Hope that helps. 😊
 
Last edited:
The Highlander said:
@khansaheb,
Not sure why you have introduced this. I don't see how it can help to achieve the desired result. 🤔
(KISS 😉)
Click to expand...
Because during the first day of class of "mechanics of materials", while discussing vector dot-product a similar problem was introduced. That is why I posted further questions for clarification.
 
khansaheb said:
Because during the first day of class of "mechanics of materials", while discussing vector dot-product a similar problem was introduced. That is why I posted further questions for clarification.
Click to expand...
I see but I trust you realize that it's use isn't particularly helpful here as you could only use the dot-product of the applied force and resulting displacement vectors to find the work done in moving the block if there were no frictional force involved. (Then, 5 × 20 × cos 60° = 50 would give you the (total) work done (as 50 J) always assuming, ofc, that the Mass of the block was greater than c.1¾ kg which would mean that there was no increase in Potential Energy involved.)

But since the Frictional force of 4 N in the direction opposite to the displacement is stipulated and, therefore, has to be taken into account, this means that one first has to find the horizontal component of the applied force so that the resultant force in the horizontal direction may be determined.

Of course, once you have found that vector quantity, you could then use the dot-product of that (6 N) resultant force and the (5 m) displacement to find the work done (only) in the horizontal direction but, since the angle between these vectors is (now) 180°, then cos θ becomes 1 and the calculation, therefore, simply reduces to Work Done = Force × Distance again (ie: 30 J).

I could see where you were heading with your questions (and accept that there is a place for that kind of Vector Multiplication in certain circumstances) but my concern was that, given that the OP appeared to be having difficulty understanding the underlying Physics (or Mechanics if you prefer) in this problem, I felt it might only serve to confuse him/her even further, especially since it didn't, in this case, offer any more elegant or efficient approach to reaching the desired solution.

I trust you will, therefore, understand my motives in 'questioning' your introduction of the Vector Multiplication 'topic' and accept that there was no offence intended in my doing so.

Regards,
B. 😊
 
Last edited:
khansaheb said:
That is why the reference to "FBD" is made
Click to expand...
I'm sorry, what "reference to "FBD" is made"??? 🤷‍♂️

Have I missed something? I don't see a reference to "FBD" anywhere in the thread; not that it would change my point (about keeping things simple) even it were mentioned. 🤔
 
You must log in or register to reply here.
Top