Block hitting a spring down an incline

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SUMMARY

The discussion revolves around a physics problem involving a block of mass 19 kg released on a frictionless incline at an angle of 34°. The block compresses a spring by 3.6 cm before stopping. The correct distance moved down the incline is calculated as 0.290429321 meters. However, the calculation for the block's speed upon touching the spring is incorrect, as the user misapplies the formula for kinetic energy and spring compression.

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AnkhUNC
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Homework Statement


In Fig, a block of mass m = 19 kg is released from rest on a frictionless incline of angle θ = 34°. Below the block is a spring that can be compressed 4.5 cm by a force of 210 N. The block momentarily stops when it compresses the spring by 3.6 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c08/fig08_41.gif


Homework Equations





The Attempt at a Solution



I found (a) = 0290429321 meters which is correct.
For (b) I tried the following v = sqrt(K*.036m^2)/19kg. But this is also incorrect. Any ideas?
 
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AnkhUNC said:
I found (a) = 0290429321 meters which is correct.
For (b) I tried the following v = sqrt(K*.036m^2)/19kg. But this is also incorrect. Any ideas?

No decimal point for (a)?

For (b), your approach is correct, but I'm unable to make out if the sqrt covers the 19 in the denominator.
 

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