- #1
kopinator
- 41
- 1
The cable of an elevator of mass M = 1640 kg snaps when the elevator is at rest at one of the floors of a skyscraper. At this point the elevator is a distance d = 36.4 m above a cushioning spring whose spring constant is k = 13600 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 12724 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed. (1/2)kx^2=Spring energy
K=(1/2)mv^2
U=mgh
Vf^2=Vi^2+2a(X-Xi)F=ma
mg-Ff=ma
a=2.05 m/s^2
Vf^2=Vi^2+2a(X-Xi)
Vf^2=0+2(-2.05)(-36.4)
Vf=12.22 m/s
(1/2)kx^2=(1/2)mv^2
(1/2)(13600)x^2=(1/2)(1640)(12.22^2)
x= 4.24 m (that didn't work)
(then I tried this)
(1/2)kx^2=(1/2)mv^2+mgh
x=10.20 m (still wasn't correct)
K=(1/2)mv^2
U=mgh
Vf^2=Vi^2+2a(X-Xi)F=ma
mg-Ff=ma
a=2.05 m/s^2
Vf^2=Vi^2+2a(X-Xi)
Vf^2=0+2(-2.05)(-36.4)
Vf=12.22 m/s
(1/2)kx^2=(1/2)mv^2
(1/2)(13600)x^2=(1/2)(1640)(12.22^2)
x= 4.24 m (that didn't work)
(then I tried this)
(1/2)kx^2=(1/2)mv^2+mgh
x=10.20 m (still wasn't correct)