## Elastic String,Beam and Object

1. The problem statement, all variables and given/known data

A mass of 10kg is attached to one end of an elastic string of unstretched length 1.5 m, the other end of the string being fixed to a beam located 2.5 m above ground level. The spring constant of the string is 600 Nm-1. If the mass is allowed to fall freely from the beam, at what distance above the ground will it first come to rest? (for this question, assume that g = 10 m s-2.)

2. Relevant equations

U=mgh
E=1/2 D*A2

3. The attempt at a solution

I am really confused here.

First of all , i found by how much the wire is stretched when the object is attached .so
ΣF=0 <=> mg=kx <=> x= (1/6) m .

When we lift the object to the height of the beam the object will have Potential energy =mgh = 10*10*1.5=150J ( from the equilibrium position )

So i said that 1/2 D*A2 = 150 so A=0.707 m

The distance of the position of equilibrium from the floor will be 2.5 - 1.5 -(1/6)=5/6 .

So the object will go as down as 5/6 + A Therefore the minimum distance from the floor will be (5/6) - A = 0.1263 m .

is my solution correct ? Thank you for your time :°)
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