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    wdwinell's Avatar
    wdwinell Posts: 1, Reputation: 1
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    #1

    Jun 11, 2007, 09:41 PM
    Calculate work done
    I was tutoring an 8th grade student in Physical Science and came across this question:

    Two students apply 1,000 N of force jointly to move a 3,000 N car 10 meters in 100 seconds.

    How much work was done?

    How much power was applied?

    I know that W = F d

    My question is: Does the force include the weight of the car?

    I thought the answer was 10,000 joules (1,000 *10) and 100 watts but do I include the weight of the car in calculating the total work done?

    Thanks'

    Bill
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Jun 11, 2007, 11:54 PM
    No, I believe you are correct. The way the question is worded (including the car's weight in newtons) is obviously testing understading of the difference between force and weight when calculating work.

    As I can't tell if you've studied physics in any depth, I will assume you might have. The full equation is actually W = F.d where . Is the vector dot product and bold type represents a vector. This introduces a factor depending on the direction of the distance and the direction of the force. Since the distance moved is perpendicular to the weight vector, this factor is 0 for the weight, and so contribution to the work by the weight is 0, similarly, because the students are pushing it in the same direction as it is moving, the factor for them is 1. So they fully contribute to the work.

    I hope I haven't confused you by introducing a mathematical concept that you are not familiar with. Please ask further... But I think I have at least reassured you that the weight should not be factored in since the car moves no distance in the direction of the weight vector.
    mmjdny's Avatar
    mmjdny Posts: 13, Reputation: 1
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    #3

    Jun 13, 2007, 09:27 AM
    work done = applied force X distance moved

    av. Power = WD/ total time
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Jun 13, 2007, 09:30 AM
    Hi mmjdny, welcome to ask me helpdesk. Please read questions before you answer them, as you have not addressed the question that wdwinell asks at all.
    dayslug24's Avatar
    dayslug24 Posts: 62, Reputation: 5
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    #5

    Jun 14, 2007, 09:24 AM
    work =30,000
    power=100

    to calculate the force used, you need the weight of the car, multiplied by the distance it was moved.
    Divide that answer by the time it took to move, and we get how much power was used.
    dayslug24's Avatar
    dayslug24 Posts: 62, Reputation: 5
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    #6

    Jun 15, 2007, 06:09 AM
    Hmmm... I guess I should brush up on my basic physics... :(
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #7

    Jun 15, 2007, 09:12 AM
    That's okay, we're all here to learn and the best way to do that is through making mistakes. The only thing the disagree is for is to make sure that nobody thinks that it is right, it isn't a reflection on you or anything, as some people seem to think it is.
    hayjay1395's Avatar
    hayjay1395 Posts: 1, Reputation: 1
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    #8

    Nov 12, 2008, 07:29 PM

    I'm in 8th grade science right now, and I do think you include the weight of the car with the work done and the power. But I'm not sure if this is totally right. :confused:
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #9

    Nov 13, 2008, 01:39 AM

    Nope. You don't include the car's weight. The car's weight is not taken into consideration, since the only force applied by the two students in the question is 1000 N. The 3000 N are due to gravity, and by Newton's third law, the ground already creates a reaction force of 3000 N.

    If you were lifting the car (only lifting), then, you'll apply the weight of the car.
    harum's Avatar
    harum Posts: 339, Reputation: 27
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    #10

    Nov 13, 2008, 10:27 PM
    Well, the real answer is we do not know the direction of the force applied to the car and the direction of the displacement due to this force, and the good thing is that we do not need to know this. Because we are only asked to calculate the work done by the students. This work is the product of the force exerted on the car by the students and displacement in the direction of that force. Strictly speaking, from the text we do not know if the directions of the force and displacements coincide. The easiest thing is to assume that they do. It well may be that this problem is accompanied by a drawing, which clarifies the matter. It is possible that the students pull this car up the ramp, in which case the work is done against the gravity of the car too. What we can say for sure is that the students do not lift the car, as their force is not sufficient for this. But what is important here is that students only applied a certain force (1000N) and it is this force that is used to calculate the work done by students. To put it in other words, your work is only measured by the force that you apply, which is 1000N here. You could apply the same force to a car that is heavier, but your work would still be the same provided your force and displacements are the same. HTH
    davey7282's Avatar
    davey7282 Posts: 1, Reputation: 1
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    #11

    Sep 13, 2010, 10:27 AM
    force x distance moved in the direction of the force = work done...

    1,000N x 10m = 10,000J (or 10kJ)

    applied power = work / time...

    10,000J / 100s = 100 watts
    MissSavage's Avatar
    MissSavage Posts: 1, Reputation: 1
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    #12

    May 9, 2012, 05:38 PM
    I teach grade 8 science. The teacher threw in the 3,000N to confuse the students. It is superfluous information. To calculate the work you use the formula:

    Work(J)=Force(N) x Distance(m)
    Work=1000Nx10m
    Work=10,000J

    The 100 seconds was added to the question to let them answer question number 2. Power is measured in watts and watts is a measure of joules per second.

    Watts=joules/second
    Watts=10,000J/100s
    Watts=100

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