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Elastic potential energy question

  • Thread starter Qred
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  • #1
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A slingshot will shoot a 10-g pebble 22.0 straight up. How much potential energy is stored in the slingshot's rubber band?

the solution is U = mgy = (10 g)(9.8 m/s^2)(22.0 m) according to the book.

does this mean the potential energy of the slingshot's rubber band is equal to the gravitational potential energy of the pebble? does it also mean the speed of the ball as it comes down is equal to its speed driven by the elastic potential energy of the slingshot's rubber band given air resistance is negligible? i don't get the solution since getting the "U sub el" requires distance from point where the rubber is stretched to the point where it is relaxed times the "k" constant and 1/2. i don't know how to get "k" and "how far it is stretched". yet the solution suggests otherwise. please enlighten me with this problem...
 
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  • #2
Andy Resnick
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Assuming there is no dissipation (i.e. the rubber band doesn't heat up, no air resistance, etc), then you are correct. The elastic energy is equal to the amount of kinetic energy imparted to the pebble, enabling it to rise 22.0 (units?).
 
  • #3
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Assuming there is no dissipation (i.e. the rubber band doesn't heat up, no air resistance, etc), then you are correct. The elastic energy is equal to the amount of kinetic energy imparted to the pebble, enabling it to rise 22.0 (units?).
what if we can get the distance from the stretched point to the relaxed point of the slingshot and the k constant is given, can we get the same answer? also, could you please answer the 2nd question in my first post? thanks
 
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  • #4
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never mind..it's crystal clear now. thanks a lot!
 

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