Spring force, find the speed at equilibrium

  • #1
A .50 kg block sliding on horizontal frictionless surface is attached to one end of a horizontal spring (with k = 500 N/M) whose other end is fixed. The block has a kinetic energy of 20J as it pass through its equilibrium position (the point at which the spring force is zero.)
  1. what is the speed of the block as it passes through its equilibrium position?
  2. at what rate is the spring doing work (power) as the block passes through the equilibrium pont?
  3. what is the maximum compression of the spring?
  4. at what rate is the spring doing work on the block as the block passes through a point 5 cm away from the sprig's equilibrium position and being stretched?


The Attempt at a Solution


1.
I know this is wrong but
do I use K - 1/mv2
20 =.5*.50*v2
v -8.9 m/s


I missed my lecture the day when his was taught, I dont have any idea how to solve any of this ... can you please help me??
 

Answers and Replies

  • #2
kuruman
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Yes that is what you use. The 20 J of energy is the total energy of the oscillator no matter where the mass is as it moves back and forth.
 
  • #3
can you give me hints on the other three questions too
 
  • #4
kuruman
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For 2 and 4, what is an expression for power that you can use that involves the speed?
For 3, how would you express the total energy at maximum compression?
 

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