The cable of the 1400 kg elevator cab in the figure snaps when the cab is at res

Question

The cable of the 1400 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 5.0 m above a spring of spring constant k = 0.24 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 2.5 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Explanation / Answer

a) Gravitational potential energy- frictional force=change in Kinetic energy

     mgh - fk = (1/2)mv^2

(1400*9.8*5)-2500=(1/2)(1400)(v^2)

66100=700(v^2)

v^2=94.428

v=9.71 m/s

b) net workdone on the spring = (1/2)Kx^2

                               66100=(1/2)*(0.24X10^6)x^2

                               x=0.7422 m or 74.22 cm

c) Net workdone by restoring force of spring = mg S+ friction force

                           66100= 1400*9.8*S + 2500

                                  S= 4.636 m

d) Networkdone = Energy lost due to friction (or workdone by friction)

           68600 = fk.S

           S = 68600/2500= 27.44 m

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