Find max compression of spring, block on incline

AI Thread Summary
A 5 kg block on a frictionless ramp at a 30-degree angle compresses a spring as it slides down. The gravitational potential energy is converted into spring potential energy, leading to the equation mgh = (0.5)kx^2. The height h must account for the distance the block travels down the incline, which is influenced by the spring's compression. The correct height formula is h = (1.3 + x)sin(30 degrees), indicating that the block travels more than 1.3 m down the ramp. This clarification helped resolve the confusion regarding the maximum compression of the spring.
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Homework Statement


A 5 kg block is placed near the top of a frictionless ramp, which makes an angle of 30o degrees to the horizontal. A distance d = 1.3 m away from the block is an unstretched spring with k = 3000 N/m. The block slides down the ramp and compresses the spring. Find the maximum compression of the spring.


Homework Equations


mgh = PEgravitational
(0.5)kx^2 = PEspring

The Attempt at a Solution


Since energy is conserved, i set them equal to one another, with h = sin(30)*1.3. x gives 0.146m as maximum compression. Webassign doesn't take this as the right answer. I tried all numbers from 0.01 - 1.99, none are the right answers. This question is making me want to smash my computer. (Also, I tried negative values too.)
 
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Your method and your answer is correct. Does Webassign ask for it in cm or another unit?
 
rock.freak667 said:
Your method and your answer is correct. Does Webassign ask for it in cm or another unit?

It's in meters. If the answer is correct, I'd check with the instructor, thanks.
 
If the spring is compressed by an amount x, then h is given by:

h=(1.3+x)\sin(30^{\circ})\,.

The block moves more than 1.3m down the incline.
 
SammyS said:
If the spring is compressed by an amount x, then h is given by:

h=(1.3+x)\sin(30^{\circ})\,.

The block moves more than 1.3m down the incline.

oh wow, that makes sense now, thanks!
 
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