the elevator is traveling at 30m/s just as it touches the spring. The safety clamps engage at that moment and provide 20000 N of frictional force. the mass of the elevator is 3000 kg and the spring constant is 1.5x10^5 N/m. how fast will the elevator be traveling after the spring compresses .8m?

Question

anonymous · Answer

Kinetic energy of the elevator at contact is 1/2mv2 = 1/2 . 3000 . 30 . 30 = 1.35MJ
Work done by friction = 0.8 . 20000 = 16kJ
Energy stored in spring is 1/2 kx2 = 1/2 . 1.5E5 . 0.8 ^2 = 48 kJ
Gravitational energy gained = mgh = 3000 . 9.81 . 0.8 = 23.544 kJ
Kinetic Energy at 0.8m = 1350kJ-16kJ-48kJ+23.544kJ = 1309.544kJ
V = (2 . E / m) ^ 0.5 = 29.54 m/s

anonymous · Answer

thank you!

anonymous · Answer

what does E stand for in the last part of the problem, im not getting that answer

anonymous · Answer

E is the Kinetic Energy that is found from the initial Kinetic energy and all the changes to the energy from friction, spring and gravity. Rearranging E = 1/2mv2 gives E = sqrt(2E/m)

anonymous · Answer

so what value did u use fr E?/

anonymous · Answer

1309.544 kJ :)

anonymous · Answer

Note the kJ i.e. 1.309544 E6 J

anonymous · Answer

thats where my problem was thank you again!