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posted by  Jfcap on 10/11/2008 1:31:16 PM  |  status: Live  

Hooke's Law 7.50

Course Textbook Chapter Problem
Algebra Based Physics N/A N/A N/A
Question Details:
A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 2.00 {\rm cm} against the elastic band, the pebble goes 3.00 {\rm m} high.

Assuming that air drag is negligible, how high will the pebble go if you pull it back 3.00 {\rm cm} instead?
 

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posted by Justin_08 on 10/11/2008 3:41:34 PM  |  status: Live
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Jfcap's comment:
"thanks"
Response Details:
 
 
             Given that the intial height is h = 3.00 m
             The pulling length is  x1 = 2.00 cm = 2.00*10-2m
              The pulling length is  x2 = 3.00 cm = 3.00*10-2
 -------------------------------------------------------------------------
  ( if the bocy is in compressed position it has eleastic potential energy and at height h it has gravitational potential energy)  
 Apply conservation fo energy  then we ger
                             (1/2)kx12 = mgh
            where  g = 9.8 m/s2 is  the accelaration due to gravity
 
                then we get  force constant  k = ----------- N/m 
 
                            (1/2)kx22 = mgh2
              then we get h2 = -------- m
                                     
            
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