A spring 1.50m long with force constant 475 M/m is hung from the ceiling of an e

Question

A spring 1.50m long with force constant 475 M/m is hung from the ceiling of an elevator, and a block of mass 10.0 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? B) If the elevator subsequently accelerates upward at 2.00 m/s^2, 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. 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?

*Please draw, write equations, and provide explanation for each part.

Explanation / Answer

a) weight = 10.0 * 9.81 N
when the block is slowly lowered to its equilibrium point, the spring stretched
x = 10.0 *9.81 / 475
= 0.206 m

(b) If the elevator subsequently accelerates upward at 2.0 m/s2,
additional force = mass * acc
= 10.0 * 2.0
= 20.0N
position of block from y=0
= 20.0/ 475
= 0.0421 m

(c) If the elevator cable snaps during the acceleration, subsequent motion of the block relative to the freely falling elevator is SHM.

amplitude of its motion = 0.0421 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.