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Work and energy (Using calculus to solve)

  1. Nov 19, 2003 #1
    Someone plz help me to do these problems below...show me what foruma/ integral to use... Thank you very much.
    1) A cable 600 ft long that weighs 4 lb/ft is hanging from a windlass. How much work is done in winding it up?
    2) A 5-lbs monkey is attached to the end of a 30-ft hanging chain that weighs 0.2 lb/ft. It climbs the chain to the top. How much work does it do?
    3) If R is the radius of the earth(about 4000 mi) and g is the acceleration due to gravity at the surface of the earth, then the force of attraction exerted by the earth on a body of mass m is F=mgR^2/r^2, where r is the distance from m to the center of the earth. If this body weighs 100lb at the surface of the earth, what does it weigh at an altitude of 1000mi? How much work is required to lift it fromt eh surface to an altitude of 1000 mi?
  2. jcsd
  3. Nov 19, 2003 #2
    #1: ΣW = ∫F dx = m∫a dx = m∫dv/dt * v*dt = m∫v dv. Since it's a definate integral from 0 to 600, it = 4[(600)^2/2] - 0.

    &Simga;W = 120000J.
  4. Nov 19, 2003 #3
    #3: You sure it's F = mgR^2/r^2? Not the universal gravitation equation? m,g,R,r are all constants... and none of them can be integrated

    Unless you use I = ∫r^2 dm...
  5. Nov 20, 2003 #4


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    I'm going to take a swing at this one. I will put the force as this:
    F = 5 + (0.2)x/2
    I'm thinking this is the formula because the weight of the monkey is a constant 5 and the length of the chain is only half of the height he climbs. If he climbs up the full 30ft , only 15ft of chain will hang from him.
    F = 5 + 0.1x

    E = รง5 + 0.1x dx
    E = 5x + 0.1x^2/2
    E = 5x + 0.05x^2
    E = 5(30) + 0.05(30)^2
    E = 195ft-lbs

    I don't know the imperial system so I don't know what your unit for energy is.
  6. Nov 20, 2003 #5

    Doc Al

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    Staff: Mentor

    Re: Re: Work and energy (Using calculus to solve)

    Huh? Yes, you got the "right answer" but who knows how. :smile:

    Try this:
    W = ∫F ds , where ds is the displacement.
    The force at any point is the weight of the cable that's left hanging. Call the hanging length "x". (xi= 600; xf= 0) Thus:
    F = λ x , where λ is the weight per unit length.
    How does ds relate to dx? As ds goes up, dx goes down (the hanging length gets shorter): so, ds = -dx. Thus,
    W = ∫F ds = - λ∫x dx (from xi to xf) = λ∫x dx (from xf to xi) = λxi2/2

    Note to gigi9: Give the problems a try before asking for help! Then the help would make more sense. (Also, maybe you'll get more help in the Homework Help section.)
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