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    Elastic strong: Lambda = 12, l = 0.5, particle attached to end of mass 0.5, other end attached to P. Particle pulled down until 1.5m below P.

    Particle is released:

    1.) Show KE = 7.1J when string slack.
    2.) Find KE when 0.5 above P
    3.) Find max h above P

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    (1)
    When the string is extended by x, it has a PE of (1/2)lambda(x/l) x = 12x^2.

    KE when string first becomes slack
    = string's initial PE - 0.5g
    = 12 - 4.9
    = 7.1J

    (2)
    When the particle is 0.5m above P, it is 1m above the point at which the string became slack.

    KE at that moment
    = 7.1 - 0.5g
    = 2.2J

    (3)
    7.1/(0.5g) - 0.5
    = 0.95m
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    (Original post by Jonny W)
    (1)
    When the string is extended by x, it has a PE of (1/2)lambda(x/l) x = 12x^2.

    KE when string first becomes slack
    = string's initial PE - 0.5g
    = 12 - 4.9
    = 7.1J

    (2)
    When the particle is 0.5m above P, it is 1m above the point at which the string became slack.

    KE at that moment
    = 7.1 - 0.5g
    = 2.2J

    (3)
    7.1/(0.5g) - 0.5
    = 0.95m
    I get h in (3) to be 0.770m above P :confused: could you explain how you got 0.95m.

    Also, to the thread starter, do you have the answers?
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    (Original post by Jonny W)
    (1)
    When the string is extended by x, it has a PE of (1/2)lambda(x/l) x = 12x^2.

    KE when string first becomes slack
    = string's initial PE - 0.5g
    = 12 - 4.9
    = 7.1J

    (2)
    When the particle is 0.5m above P, it is 1m above the point at which the string became slack.

    KE at that moment
    = 7.1 - 0.5g
    = 2.2J
    Sorry but I really don't get any of this at all!
    1.) When exactly is the string slack, what does slack really mean?
    2.) Why do you keep deducting mg when we're delaing with energy considerations?
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    (Original post by Nima)
    Johnny's first two are right, but his 3rd is wrong, its 0.770m as u said
    Sorry, I forgot that the string becomes taut again when the particle rises past 0.5m above P.

    The answer is 0.5 + x where x > 0 is such that 12x^2 + 0.5gx = 2.2. Solving the quadratic gives x = -0.68, 0.27.
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    (Original post by Nima)
    Sorry but I really don't get any of this at all!
    1.) When exactly is the string slack, what does slack really mean?
    2.) Why do you keep deducting mg when we're delaing with energy considerations?
    The string is slack when the particle is within 0.5m of P (and is taut otherwise).

    The energy equation is

    (KE of particle) + (PE stored in string) + (gravitational PE of particle) = constant

    In Part (1),

    (PE stored in string) goes down by 12
    (gravitational PE of particle) goes up by 0.5g
    so (KE of particle) goes up by 12 - 0.5g
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    (Original post by Nima)
    btw: KE increases by 7.1J - Does this mean when pulled to 1.5m below and let go, initial KE = 0? It must do for your method to be right.
    "Released" means "released from rest", so yes the initial KE is 0.
 
 
 
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