Force Spring Compression of 13kg Block on 35° Incline

AI Thread Summary
A 13.0 kg block slides down a frictionless 35° incline and compresses a spring with a spring constant of 2.60 x 10^4 N/m after traveling 3.00 m. To find the spring's compression, energy conservation principles can be applied, equating the gravitational potential energy lost by the block to the elastic potential energy stored in the spring. The weight of the block is calculated as 127.4 N, but only the component of this force along the incline affects the spring. The correct approach involves using the energy equation to determine the compression, considering the gravitational potential energy at the start and the energy stored in the spring at maximum compression. Understanding these energy transformations is crucial for solving the problem effectively.
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Homework Statement


A block of mass 13.0 kg slides from rest down a frictionless 35.0° incline and is stopped by a strong spring with k = 2.60 x 10^4 N/m. The block slides 3.00 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?


Homework Equations



F=mg (weight of block in Newtons)


The Attempt at a Solution


I've found the weight of the block in Newtons (127.4 N), and I know that I have to find the force of the spring in its direction. To get the final answer, I'd have to divide that by the value of k. But I don't know how to find the force of the spring!
 
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would that be momentary rest as the spring compresses to its maximum, before shooting the block back up, or the rest after all oscillations have died down? The former I guess.

Have you ried using energy considerations instead...consider the enrgy changes in the block and the energy stored in the spring.
 
The whole weight of the block is not pushing against the spring – just its component along the plane.

The block had only grav PE when it started. Where is that gone now? Take x as the compression in the spring and write the proper energy eqn.

The force exerted by the spring is kx.
 
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