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
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