http://www.woodcarvingguru.com/spring-clamps-2/
What is the speed of the cab just before it hits the spring?
The cable of an 1800 kg elevator cab snaps when it is at rest at a height of 3.7m above a spring constant k=1.50×10 to the 5th power. a safety device immediately clamps the cab against the guard rails producing a constant frictional force of 4.4×10 to the 3rd power opposing the cabs motion. 1. calc speed of cab just before it hits spring. 2. calc max compression produced in the spring
F = weight – friction = 1800g – 4400 = 18000 – 4400 = 13600 N
i)
F = ma
a = 13600/1800 = 7.56 m/s^2
2as = v^2 – u^2
v = sqrt (2as) as u=0
v = sqrt (2*7.56*3.7) = 7.48 m/s
ii)Best way is to use energy argument:
Initial K.E of Cab = Final P.E of Spring
0.5mv^2 = 0.5*k*x^2 where x=compression
x= sqrt(mv^2/k) = 0.819 m
So the spring will compress by approx. 820 cm