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Thread: PHYSICS help!

  1. #1
    Junior Member
    Join Date
    Oct 2004
    Posts
    29

    PHYSICS help!

    I got 2 questions which I did my work, but want to double check with someone, anyone!

    Q:A rescue helicoper lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70m/s^2 and is lifted from rest through a distance of 11m. a)what is the tension in the cable? How much work is done by b)the tension in the cable and c)the person's weight? d) use the work-energy theorem and find the final speed of the person.

    My work so far:

    a)Tension in cable

    sum of Fy=T-W=ma

    T=ma+W
    79kg(0.70m/s^2)+79(9.81m/s^2)= 830.29 N

    b)Work by tension in cable:
    W=Fcos(0 degrees)s
    =830N(1)(11m)
    =9130J

    c)work done by person's weight

    W=Fcos(180)s
    =774.99N(-1)(11m)
    =-8525 J

    d)Final speed

    KEf=W+KEo
    1/2 mv^2=W+1/2mv^2

    Vf=square root (2*9133J/79kg)
    =15.20m/s

    Any suggesions great.

    Here's the other one...

    Q:A particle, starting from point A in the darwing, is projected down the curved runway. upon leaving the runway at point B, the partticle is traveling straight upward reaches a height of 4.00m above the floor before falling back down. ignoring firction and air resistance, find the speed of the particle at point A.

    I've attached a crude paint drawing.

    Ok my work:

    Wnc (non conservative work) = 0

    Ef=Eo

    1/2mv^2 + mgh= 1/2mVo^2+mgh

    Since kinetic and potential must be constant then I can cancel some terms

    so mgh=1/2mVo^2

    Vo=square root(2*9.81*(4m-3m))
    Vo= 4.43m/s

    Sounds too simple if you ask me.

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