James S.
asked • 01/10/21Conservation of Energy Problem
A spring with a constant K = 200 N/m is attached to a wall horizontally. It is stretched a distance of x = 0.25 m. The block is released and allowed to move. Assume surface is frictionless, and block-spring is at ground level.
Find the velocity of the block at the moment it reaches the springs equilibrium point.
Please help. These conservation of energy problems are really hard for me to understand.
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1 Expert Answer
You can thing in terms of absolute energy in each state or change in energy between the two states. In this cases the only energies of interest are kinetic and spring potential. I'll start with the full energy equation:
ΔEk + ΔEP + ΔEspr +ΔEother = Won + -WFric + Qin
This is for all situations (You could include chemical, nuclear, phase changes, and, most importantly, ΔU = mcΔT). In typical Physics problems, you only include the first three mechanical energies and heat is considered not to be a factor (Q=0).
In this case, there's no W and no Wf and no ΔEp (no change in height), therefore,
ΔEk +ΔEspr = 0 or Ek,a + Espr,a = Ek,b + Espr,b where a means after and b means before the event.
Before, Ek = 0 and Espr = 1/2 kx2 and after Ek = 1/2 mv2 and Espr = 0
Plug in and solve for v. Good luck! I hope that helped.
PS. In terms of thinking through the problem quickly: You know that the initial spring energy must turn into kinetic energy. You can just equate the two as the total energy is constant (This happens when W=0, Wf=0 and Q=0)
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Michael C.
The total energy of the system has two components: Potential energy (PE) stored in the spring, and Kinetic energy (KE) of the moving mass. In the absence of friction, the sum of these two - the total energy - remains constant. When the spring is stretched but not let released, the KE = (1/2)(M)(V*V) is zero (since V=0), but the PE = (1/2)(K)(X*X) = (1/2)(200 N/m)(0.25m*0.25m) = 6.25 J. At the moment when the mass swings past the equilibrium point of the spring, the spring's displacement X (and therefore PE) is zero but because the mass is moving the KE is not. Since the total energy is constant, the KE at this moment must also be 6.25 J. If we know the mass M, we can solve for the velocity V. I don't see the mass listed as the problem is presented, but this is the approach.01/10/21