Recent content by tobias.mueller2

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    Work - Acceleration of an elevator

    w=∆e[kyn]+∆e[pot]+637,65 N*1.5 m=0.5*m*v2-0+m*g*s-0+637,65 N*1.5 m=0.5*1300 kg*1.5 m/s2+1300 kg*9.81 m/s2*1.5 m+637,65 N*1.5 m=21548,475 Joule In the end, I get the same result. Do I made any fault?
  2. T

    Work - Acceleration of an elevator

    Homework Statement An elevator with 1300 kg accrlerate from 0 m*s^-1 to 1.5 m*s^-1 in 2 s. After that, he drives further with a constant velocity of 1.5 m*s^-1. One must add a frictional force of 637,65 N. The questions are: 1. How much work is needed to accelerate the elevator? 2. How...
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