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An elevator and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 12.0 m/s, is brought to rest with constant acceleration in a distance of 42.0 m.

Hrishant Goswami , 10 Years ago
Grade 10
anser 2 Answers
Jitender Pal

Last Activity: 10 Years ago

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The negative sign in the acceleration indicates that the elevator is decelerating.
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Therefore the tension in the supporting cable is 18,438.4 N .
Kushagra Madhukar

Last Activity: 5 Years ago

Dear student,
Please find the solution to the problem.
 
Sign convention → Let the downward acceleration and velocity to be positive and the upward to be negative
We have, v2 – u2 = 2as
Hence, a = (02 – 122)/2 x 42 = – 1.714 m/s2
Now, mg – T = ma
Hence, T = m(g – a)
              = 1600(9.81 – ( – 1.714))
              = 1600 x 11.524
              = 18438.86 N = 18.44 kN
 
Thanks and regards,
Kushagra
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