
#1
May1612, 05:47 PM

P: 64

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 Nm1. 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 s2.) 2. Relevant equations U=mgh E=1/2 D*A^{2} 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*A^{2} = 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 :°) 



#2
May1612, 05:56 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

The object will sail through the equilibrium position at high speed, and come to rest well below it! Start again. 



#3
May1612, 06:02 PM

P: 64

Yeah that's what i am saying . It will come to rest every time it reaches +A or A , right ?



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