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posted by  CHMSTRY on 11/6/2009 11:17:45 PM  |  status: Closed  |  Earned Karma: 55

lifesaver

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/10/2009 at 7:00:00 PM
Question Details:
Stone Rests on Spring Figure 10-34 shows an 8.00 kg stone resting on a spring. The spring is compressed 9.0 cm by the stone.


Figure 10-34

(a) What is the spring constant?
1 N/cm
(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
2 J
(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
3 J
(d) What is that maximum height, measured from the release point?
4 m
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Oracle
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posted by zsm28 on 11/6/2009 11:32:15 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
m = 8.00 kg, x = 9.0 cm, d = 30.0 cm
a) k = mg/x = 8.71 N/cm
b) E = k(x + d)2/2 = 66.2 J
c) E = 66.2 J
d) mgh, so h = /mg = 0.845 m
(SME)
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posted by Kevin_08 on 11/6/2009 11:35:48 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:

Given that
 m = 8.00 kg
  x =0.09 m
 g =9.8 m/s2

 

a)  We know that
    weight of the stone is
     mg = kx
        k = mg/x
      
b)  At x' = 30.0 cm,
   The elastic potential energy of the compressed spring just before that release   is
            Ee = k(x' + x)2/2
           
c)
   From the law of energy conservation is
       Eg= Ee
 
d) We  know that
           Eg= mgh
∴  The maximum height is
               h = Eg/(mg)
 Substitute the values.
I hope this helps! Best of luck with the rest of your coursework.
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