Question

A spring 1.50 m long with force constant 448 N/m is hung from the ceiling of an elevator, and a block of mass 10.9 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.) m
(b) If the elevator subsequently accelerates upward at 1.89 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.) m
(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?

Answer 1

Answer:

Explanation:

Let the extension in the spring be x .

restoring force = weight of block

kx = mg

x = $\frac{10.9\times9.8}{448}$

= 23.84 cm

b )

When the elevator is going upwards

Restoring force = mg + ma

k x₁ = 10.9 ( 9.8 + 1.89 )

x₁ = 28.44 cm

( y coordinate will  be - ( 28.44 - 23.84 ) = - 4.6 cm )

c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0

Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to

.2844 x 448

= 127.41 N

So a net acceleration a will act on the block

a = 127.41 / 10.9

= 11.68 m / s²

The block will undergo SHM with amplitude equal to 28.44 cm .