Potential energy incline problem

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SUMMARY

The discussion centers on a physics problem involving two blocks connected by a string over a frictionless pulley, with one block attached to a spring. The 20.0 kg block is pulled down an incline and released, while the 30.0 kg block is raised above the floor. The key equation used is the conservation of energy: Ki + Ui = Kf + Uf. The participants clarify that the speed of the blocks should be calculated just before the spring is compressed, indicating that spring potential energy does not factor into this specific moment.

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  • Understanding of Newton's laws of motion
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  • Knowledge of spring mechanics, specifically Hooke's Law
  • Ability to apply conservation of energy principles in physics problems
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Homework Statement


A 20.0 kg block is connected to a 30.0 kg block by a 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 300 N/m, as shown in Figure P7.53. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0 kg block is pulled 18.0 cm down the incline (so that the 30.0 kg block is 38.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0 kg block is 20.0 cm above the floor (that is, when the spring is unstretched).
http://www.webassign.net/sj/p7-45.gif"

Homework Equations


Ki + Ui = Kf +Uf

The Attempt at a Solution


0 + (30)g(.18m) = 1/2(M + m) v^2 + (20)(g)(.18)(sin40)

I solved for v but the answer was not correct. I am not sure what I did wrong. I did not include any spring potential force because it was asking for the speed right when the 30kg block hits the spring so the spring has not yet been compressed so there is no spring potential energy.
 
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In your attempt, you have not used spring constant.
What are the forces acting on the 30 kg block is the stretched position?
What are the forces acting on the 20 kg during that time?
 
For me, the problem can be interpreted in two ways.
1. Calculating the speed right when it touches the spring so there is no spring compression yet.
2. the block compresses the spring resulting in a spring potential force.
Okay I understand now thank you rl.bhat
 

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