How Far is the Spring Compressed When a Block Slides Down an Incline?

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A block of mass 12.0 kg slides down a frictionless 32.0° incline and compresses a spring with a spring constant of 1.00 x 10^4 N/m after traveling 3.00 m. The problem is solved using the conservation of energy principle, equating initial potential energy to kinetic energy and spring potential energy. The equations used include PE = mgh and KE = 1/2 mv². After calculations, the spring is found to be compressed by 0.19 m when the block comes to rest. The solution effectively demonstrates the application of energy conservation in a frictionless system.
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[SOLVED] Please help- spring problem

1. A block of mass 12.0 kg slides from rest down a frictionless 32.0° incline and is stopped by a strong spring with k = 1.00 104 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?


2. K=F/x - K= spring constant



3. I really have no idea how to do this
 
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Do you know conservation of energy?
 
sort of PE=KE? is that what i need to use since there is no friction?
 
well, we have the initial Potential energy
Then, at the point of impact, we have less potential, because now some of that is kinetic.
Then, there is spring potential which counteracts all of the previous potential/kinetic energy.
So, express that algebraically.
 
so do i have to subtract potential energy from the final kinetic energy?
 
i just got it.
PE=mgh=KE=1/2mv2
then i plugged my numbers in and got x to be .19 and it was right
thanks
 
Last edited:
OK good
 
Last edited:

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