Energy Due to a Spring? Please Help!?

Question

An elevator cable breaks when a 795 kg elevator is 26.3 m above the top of a huge spring (k = 8.00 104 N/m) at the bottom of the shaft.

(a) Calculate the work done by gravity on the elevator before it hits the spring.

(b) Calculate the speed of the elevator just before striking the spring.

(c) Calculate the amount the spring compresses (note that here work is done by both the spring and gravity).

Answer 1

a)Potential energy equal to work done.
Pe=W=mgh
Pe= 795 x 9.81x 26.3= 20.5 E +4 J

b) Potential energy gets converted to kinetic energy
Ke=Pe
0.5mV^2=mgh
V= sqrt(2gh)
V=sqrt(2 x 9.81 x 26.3)=22.7 m/s

c) Potential energy of the elevator equal to energy stored in the spring
Pe=Ps
Ps=0.5 k x^2
mgh=0.5 k x^2
x=sqrt(2mgh/k)
x=sqrt(2 x795 x 9.81x 26.3/8.00 E+4 )
x=2.26 m

Answer 2

a) work done by gravity = m x g x h
= 795 x 10 x 26.3 = 209085 joule
b) this work done by gravity converts in to kinetic energy of elevator there for, 1/2 m v^2 = mgh
v = 22.934 m/s.
c) let the compression in spring be x
m x g x ( h + x ) = 1/2 k x ^2
x=2000.564m