How Far Will the Spring Compress When a Box Falls on It?

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The problem involves a 2.14 kg block dropped from a height of 43.6 cm onto a spring with a force constant of 18.6 N/cm. The initial velocity of the block upon impact is calculated to be 2.92 m/s. The energy conservation equation is applied, but the user mistakenly assumes the final velocity and height are zero, leading to an incorrect calculation of spring compression. The correct approach emphasizes using energy conservation throughout and recognizing that the total drop distance includes both the height of the drop and the spring compression. The user is encouraged to revise their calculations with these considerations for an accurate result.
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Homework Statement


A 2.14 kg block is dropped from a height of 43.6 cm onto a spring of force constant 18.6 N/cm. Find the maximum distance the spring will be compressed.

I know that the mass is 2.14 kg, the force constant of the spring is K=1860 N/m.

I assume the final velocity, the final height of the block after being dropped, and the final distance the spring will be compressed are 0.


Homework Equations


U+K=0
v^2=v(0)^2+2AS


The Attempt at a Solution


To find the initial velocity, I use the constant acceleration equation:

0=V(0)^2-19.7(.436); V(0)=2.92m/s

Since I know (or assume) a conservative force, I use the equation:

U+K=0

mgh+(1/2)mv(0)^2+(1/2)kx^2=0

2.14(-9.8)(.436)-.5(2.14)(2.92)^2+(1/2)(1860)x^2=0

x=.139 m = 13.9 cm

The book got 11.1 cm. Can anyone tell me what I did wrong and how to do the problem correctly? Must help is appreciated. Thanks!
 
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The initial (and final) velocity is zero. What you found is the speed at which the block hits the spring. That would be OK, but then you messed up the energy conservation.

It will be much easier if you just stuck with energy conservation from start to finish. Hint: the total distance dropped is not just 43.6 cm.
 
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