physics

A descending elevator of mass 670 kg is uniformly decelerated to rest over a distance of6 m by a cable in which the tension is 8683 N.The acceleration due to gravity is 9.8 m/s2.Calculate the speed vi of the elevator at the beginning of the 6 m descent. Answer in units of m/s.

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  1. The deceleration rate a, stopping distance X and the initial velocity Vi are related by
    Vi = sqrt(2 a X)

    The cable tension minus the weight is the net force

    F = T - M*g = 8683 - 6566 N = 2117 N

    Use F = ma to get a and then use the first equation to get Vi.

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  2. i got the correct answer but i still get any of the concept >.< i feel ahhh oh oh any good sites to learn physics really good

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  3. If you got the correct answer, you must be learning something.

    Did you omit the word "don't" after still?

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  4. What is the time

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  5. about 7

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