Asked by christine
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.
Answers
Answered by
drwls
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.
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.
Answered by
christine
i got the correct answer but i still get any of the concept >.< i feel dumb ahhh oh oh any good sites to learn physics really good
Answered by
drwls
If you got the correct answer, you must be learning something.
Did you omit the word "don't" after still?
Did you omit the word "don't" after still?
Answered by
Anonymous
What is the time
Answered by
artor
about 7
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