Spring/Incline plane kinematics/work problem

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A 0.5 kg block is compressed against a spring with a constant of 625 N/m and then released on a frictionless incline at a 30-degree angle. The initial potential energy of the spring, calculated as (1/2)kx², converts entirely into gravitational potential energy as the block ascends the incline. The relationship between the distance traveled up the incline (d) and the height gained (h) is crucial for solving the problem. By equating the initial spring potential energy to the final gravitational potential energy (mgh), the maximum distance the block travels can be determined. This approach effectively utilizes energy conservation principles to find the solution.
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1. A .5 Kg block rests on a horizontal fricionless surface. The block is pressed back against a spring having a constant of K =625 Newtons per meter. The spring is compressed .1 meters. Then the block is released. Find the maximum distance the block travels up the frictionless incline if the angle on the incline is 30 degrees.


2.



3. I set up the PE, KE, and spring .5kxsquared added up equal total energy. For hight, i used 0, where the only stored energy would be the spring(.5kxsquared), PE and Ke here is 0. Than I put H as it's max height(where it would stop on the incline). I don't know what H is though, and can't solve for V.
 
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this has nothing to do with kinetic energy, you can calculate the height through use of the initial and final potential energy.
 
Shadowsol said:
1. A .5 Kg block rests on a horizontal fricionless surface. The block is pressed back against a spring having a constant of K =625 Newtons per meter. The spring is compressed .1 meters. Then the block is released. Find the maximum distance the block travels up the frictionless incline if the angle on the incline is 30 degrees.


You said the surface was horizontal right at the beginning...

As Oerg said, you can solve it by using the fact that the initial PE of the spring is entirely converted to grav PE.

Initial PE = (1/2)kx^2.

If the block travels a dist of d up the plane, and the vertical dist gained is h, then find the relation between d and h. I leave that to you.

Final PE = mgh.

Now you can find d.
 
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