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i guess it will be the worlds most longest method finding the mass and all
forget it -
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Biology; Chemistry; Physics: Post your doubts here!
- Thread starter scouserlfc
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i guess it will be the worlds most longest method finding the mass and all
forget it 
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A person exerts a horizontal force of 500N on a box, which also experiences a friction force of 100N. How much work is done against friction when the box moves a horizontal distance of 3 m?
Is the answer 1200 J or 300 J?
Is the answer 1200 J or 300 J?
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That is a work done question so lets put the information we have from the question in to something that we can easily understand and then derive which formula to use!
The information provided are that :-
There is a horizontal force of 500N and a Frictional Force of 100N
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and we have been told that the box is moved to a distance of 3m horizontally!
Work Done = Force x perpendicular Distance!
Which means we have to get the resultant force !
Resultant force = 500N - 100N { Friction means the opposing force } so subtraction!
Resultant force = 400N
Work Done = Force x Perpendicular distance
= 400 x 3
= 1200J
The information provided are that :-
There is a horizontal force of 500N and a Frictional Force of 100N
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and we have been told that the box is moved to a distance of 3m horizontally!
Work Done = Force x perpendicular Distance!
Which means we have to get the resultant force !
Resultant force = 500N - 100N { Friction means the opposing force } so subtraction!
Resultant force = 400N
Work Done = Force x Perpendicular distance
= 400 x 3
= 1200J
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Nah man i think KyleoAhsan is correct again because we want the work done against friction and not work done by frictional force against the box movement so its 1200 J .I will double check it because this redspot is really unpredictable and let you know soon
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You are right! Redspot does give the answer as 300!
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They have taken the frictional force instead of the horizontal force! But to be honest...since the question is asking for the work done horizontally..the question strictly states that " how much work is done against friction when the box moves a horizontal distance of 3m...so I believe that's the only possible answer for it!
Well I couldn't even find the past paper online!
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hi ..since iam new in o level i am bit confused in vectors can any one help me...
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First what is magnitude? It is a number and the S.I Unit! For example!
5 seconds = 5 s or 500 Newton = 500N
Now in a scalar quantity there is only magnitude! whereas in vectors there is magnitude AND direction!
The difference between vector and scalar is that vector has a direction given!
For example :
* A car travelling in a journey for 24 minutes at a speed of 30km/h
That is a scalar quantity! because no direction has been given!
* A car travelling east in a journey for 24 minutes at a speed of 30km/h
Now that is a vector quantity as we know the direction its travelling in the journey!
Addition and subtraction of such vectors are important too!
For example a vector maybe ---> 5m in that direction we go the same distance but backwards <--- now the vector becomes negative! It is now (-5m) as the direction has been changed!
Some example of vectors
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1. Velocity [m/s]
2. Displacement [m]
3. Acceleration [m/s²]
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You are right! Redspot does give the answer as 300!
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They have taken the frictional force instead of the horizontal force! But to be honest...since the question is asking for the work done horizontally..the question strictly states that " how much work is done against friction when the box moves a horizontal distance of 3m...so I believe that's the only possible answer for it!
Well I couldn't even find the past paper online!
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Which year is this guys may i know please
??????????
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November 2001/Paper 1/Question 9Which year is this guys may i know please
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Which year is this guys may i know please
Well I made a little mistake its actually 300J because its the work done against friction! which means how much work is done by the frictional force as the box moves a horizontal distance of 3m!
A little more detail :
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So for example : I push a car and I exert a force of 500N and it moves and as I move there is a frictional force of say 200N so the work I am doing here is actually 200N .... 500N - 200N = the wasted effort! I put an effort of 500N but the work done is simply 200N !
I hope its correct! and hope that helps and you understand what I said
! so frictional force is 100N :
and Work Done = Force x Distance { I made a mistake in the formula its not perpendicular distance! it's simply distance }
Work Done = 100N x 3m = 300J
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SO the thing we were finding out was the work done in moving the car forwards only and not the frictional force stupid
stupid
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SO the thing we were finding out was the work done in moving the car forwards only and not the frictional force
Well somewhat it don't make sense because I don't really know how to actually explain it!
and I think I actually explained it in a silly way!
!Well it seems that Work Done = F x D in the formula [ F ] is linear not the resultant force and so 500N - 100N is the resultant force!
and if you multiply the Resultant Force x distance you get the net useful work done! and not the work done against friction!
The total work done is calculated by 500N x 3 = 1500N but that's not what we want...what we want is the work done against friction!
If you simply get the resultant force and then multiply you still don't get the work done against the friction ! but you do get the useful work done!
So in order to calculate work done against friction which means it will be positive! since friction is an opposing force
it will be ( - ) negative in nature! meaning Work Done by Friction is (-100) x 3 = -300N so the work done against friction would be 100N x 3!
Well I am not sure how you would get the meaning of it! but I hope this time it helps
!
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Well somewhat it don't make sense because I don't really know how to actually explain it!
and I think I actually explained it in a silly way!
Well it seems that Work Done = F x D in the formula [ F ] is linear not the resultant force and so 500N - 100N is the resultant force!
and if you multiply the Resultant Force x distance you get the net useful work done! and not the work done against friction!
The total work done is calculated by 500N x 3 = 1500N but that's not what we want...what we want is the work done against friction!
If you simply get the resultant force and then multiply you still don't get the work done against the friction ! but you do get the useful work done!
So in order to calculate work done against friction which means it will be positive! since friction is an opposing force
it will be ( - ) negative in nature! meaning Work Done by Friction is (-100) x 3 = -300N so the work done against friction would be 100N x 3!
Well I am not sure how you would get the meaning of it! but I hope this time it helps
dont worry man we already know what we were finding before and what we should have found instead thats why i said "stupid"
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dont worry man we already know what we were finding before and what we should have found instead thats why i said "stupid"
Oh Great! I thought my explanation example was stupid..because I made it up
! It don't make sense to me even haha! Sorry for misunderstanding ;o!
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Oh Great! I thought my explanation example was stupid..because I made it up
HAHAHAH !
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So it seems all new O3 students are pretty serious about their studies. They are asking questions a month before the start of the Session. Good Job....
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^ Just to make every one remember that anaeroid Barometer is not in the syllabus
hahahahah yes they are
hahahahah yes they are
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Hey guys,
I am a private candidate and I will be giving physics in OCT/NOV. I have completed the syllabus and I have just started doing past papers. My question is will these remaining months be enough to achieve a good grade? Also I will appreciate any advice. One more thing, a friend of mine told me that papers of OCT/NOV are harder. Is it true????
I am a private candidate and I will be giving physics in OCT/NOV. I have completed the syllabus and I have just started doing past papers. My question is will these remaining months be enough to achieve a good grade? Also I will appreciate any advice. One more thing, a friend of mine told me that papers of OCT/NOV are harder. Is it true????
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To be honest...it will be hard if you are unprepared..or if the exam contains the questions from a topic you find difficulty in...so my advice is Be very thorough...go through every topic...and master them one by one...it is sometimes useful in physics to study a little bit of advanced stuffs....it may or may not help you....as different people have different ways of studying...If you want good grades in Physics...Read the book Complete Physics by Stephen Pople...and do all the questions in there...this will make sure you are ready for most basic questions...If you do a lot of MCQ questions you will be quite thorough in physics...I suggest doing IGCSE, GCSE O level papers of Physics from 1990 - 2011 this will make sure you are very familiar with all types of questions...In physics its rare to see repeated calculation questions...so make sure you know how to use formulas..and which formulas to use....for theory paper understand every concept...reading won't help it..make sure you understand what you read...and try to explain it in your own words...use scientific words as much as possible...Think of it as explaining it to a 5th grader .. so you can make sure you explained it well enough
! That's all my suggestions and advice! Nothing is hard if you are motivated to work hard