   Chapter 13, Problem 7P

Chapter
Section
Textbook Problem

A spring 1.50 m long with force constant 475 N/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/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. (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?

a)

To determine
The stretch in the spring when the block is slowly lowered to its equilibrium position.

Explanation

Given information: The length of the spring is 1.50m . The force constant of the spring is 475Nm-1 . The bloc attached to the spring is 10.0kg . The spring is hung from the ceiling of an elevator.

Explanation:

From Hooke’s law, the formula for the restoring force is,

F=kx

Here,

F is the force

k is force constant of the spring

x is the displacement of the spring

The force on the sample is equal to the force of gravity.

F=mg

Here,

m is the mass of the object

g is the acceleration due to gravity

The force applied on the sample will be exactly equal to the negative of the restoring force

b)

To determine
The position of the block when the elevator is accelerates upward at 2.00ms-2

c)

To determine
The substituent motion of the block when the elevator cable is snapped.

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