Question

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a distance d=3.7m above a spring of spring constant k=0.15MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 4.4kN opposes the cabs 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 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.)

Answer 1

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

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