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Homework Help: How to calculate work required to move an object

  1. Jun 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Eight books, each 4.3 cm thick with a mass of 1.7 kg, lie flat on a table. How much work is required to stack them one on top of another?

    2. Relevant equations

    w= Fdcos([tex]\theta[/tex]

    3. The attempt at a solution
    For the first book, there is no work required. The displacement was zero therefore no work was required. For the second book, the d=4.31 since the book must be lifted up, but I am not sure how to calculate the force and figure the angle. Don't I need to know where the books are to begin with?
  2. jcsd
  3. Jun 30, 2008 #2
    Your angle is [tex]\theta = \pi/2[/tex] here, because the force is vertical. For the force, use gravitational pull: [tex]F=mg[/tex], where [tex]g=9.8 \unitfrac[m][s^2] [/tex] is the free acceleration.
  4. Jun 30, 2008 #3
    If I understand this right, for my second book, I would have 112 J. I took [tex]\pi/[/tex]/2 = 1.57 then multiplied that by 9.8*1.7 kg and got 26.16, then multiply that by the 4.3 cm and got 112 J for my final answer. then for the third book instead of multiplying it by 4.3 I would multiply it by 8.6 cm since I will have two books under it already.
  5. Jun 30, 2008 #4
    I get a different result. Please write down your final expression in symbolic form.
  6. Jun 30, 2008 #5
    I apologize, but I am not sure what symbolic form is. Can you please explain further?
    Thank you
  7. Jun 30, 2008 #6
    Your force and displacement are in the same direction so W = F*d
  8. Jun 30, 2008 #7
    ACasey, symbolic form is a fancy name for writing as much as possible in letters. That way, initial energy is

    [tex]E_1 = 8mgh/2 = 4mgh[/tex]

    if h=4.3cm and it labels the height of center of mass. The final energy is

    [tex]E_2 = 8mg \frac{8h}{2} = 32mgh[/tex]

    The difference is then

    [tex]E_2-E_1 = 32mgh - 4mgh = 28mgh[/tex]

    This is the answer in symbolic form. Plugging in the numbers we obtain the answer in numeric form.
  9. Jun 30, 2008 #8
    I guess I was unaware of how confused I am with this subject. How is energy coming into this problem? I thought the formula that I would use is force which is F= mass*gravity then multiply that by the displacement which in this case would be 4.3 c.m. for each book. Then I multiply that by the [tex]\pi[/tex]2 and that would be the solution in joule? Where am I getting lost?
  10. Jun 30, 2008 #9
    If you multiply force times displacement, the result is energy. The angle is not required here, so forget about it. You can follow my solution, or if you insist on your way, then the displacement for the first book is [tex]h_1 = 0[/tex], for the second book [tex]h_2 = h[/tex], for the third [tex]h_3 = 2h[/tex] and so on, then you add all them and the result obtains. Don't forget to convert everything to SI units if you want to get joules in the end.
  11. Jun 30, 2008 #10
    I'm sorry, I didn't mean to offend you. I would like to do it the correct way, but I am confused on the solution. I will get it figured out.
    Thank you for your help.
  12. Jun 30, 2008 #11
    You didn't offend me. There are several ways to calculate this result, doesn't matter too much which is used.
    Basically this is just conservation of energy problem.
  13. May 10, 2011 #12
    What about ...; A force of 30N does 180J of work across the floor . If the same
    amount of work is done over a distance of 6m, how much force is doing work ?
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