Two blocks connected with spring and pulley

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SUMMARY

The discussion centers on a physics problem involving two blocks connected by a massless string over a frictionless pulley, with one block attached to a spring. The blocks have masses of m1 = 20.0 kg and m2 = 30.0 kg, and the spring has a force constant of k = 250 N/m. The problem requires calculating the speed of each block when the spring is unstretched after the 20.0 kg block is pulled down a frictionless incline of 40 degrees. The correct approach involves applying the conservation of energy principle, where initial gravitational potential energy and spring potential energy equals the final kinetic energy of the blocks.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the conservation of energy principle
  • Knowledge of spring mechanics and Hooke's Law
  • Basic trigonometry for resolving forces on an incline
NEXT STEPS
  • Study the conservation of mechanical energy in systems involving springs
  • Learn about Hooke's Law and its applications in physics problems
  • Explore inclined plane dynamics and the effects of angles on forces
  • Review problem-solving techniques for pulley systems in classical mechanics
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of energy conservation in systems involving springs and pulleys.

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Homework Statement


A block of mass m1= 20.0kg connected to a block of mass m2 =30.0 kg by a massless string that passes over a light, frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of k =250 N/m as shown in Figure P8.64. The spring is unstretched when the system is as shown and the incline is frictionless. The 20kg block is pulled a distance 20cm down the incline of angle theta = 40 degrees and released from rest find the speed of each block when the spring is again unstreched

Homework Equations


Ei =Ef

The Attempt at a Solution


Iam confused if we should have 0.5Kx2 as the energy in the system finally. This how I think the equation should be set up but I am not sure if my reasoning is right for the 1/2kx2 being there (the mass of the block will compress the string isnt.
Taking the intial point from when m2 is 20cm down and m1 is 20cm up :

M2gh + 1/2kx2 = 1/2(M1 + M2)vf2 + M1gh + 1/2kx2

Would this be correct. At the final position the spring will compress 0.784 since:
M2g = kx
X = (20)(9.8)/250
X = 0.784

So is my reasoning correct or am I having misconceptions
 
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Two problems I see:
- The final position is with the spring unstretched. So what does that tell you about the spring PE at that point?
- Realize that the masses do not have the same change in vertical position, since one is on an incline.
 

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