A spring 1.50 m long with force constant 445 N/m is hung from the ceiling of an

Question

A spring 1.50 m long with force constant 445 N/m is hung from the ceiling of an elevator, and a block of mass 14.3 kg is attached to the bottom of the spring.

(a) By how much is the spring stretched when the block is slowly lowered to its equilibrium point? (Enter the magnitude only.)
(b) If the elevator subsequently accelerates upward at 1.50 m/s2, what is the position of the block, taking the equilibrium position found in part (a) as y = 0 and upwards as the positive y-direction. (Indicate the direction with the sign of your answer.)
(c) If the elevator cable snaps during the acceleration, describe the subsequent motion of the block relative to the freely falling elevator. What is the amplitude of its motion?

Explanation / Answer

force constant is k = 445 N/m

using Hooke's law F = k*x

but force = weight = m*g = 14.3*9.8 = 140.14

then 140.14= 445*x

x = 0.314 m

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b) when the elevator accelarating upwards

Force = m*(g+a) = 14.3*(9.8+1.5) = 161.59

then 161.59 = k*x

then x= 161.59/445 = 0.363 m

C) for a freely falling elevator,

F = m*a = 14.3*1.5 = 21.45

then 21.45 = k*x

x = 21.45/k = 21.45/445 = 0.0482 m

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