Determine the distance that the spring is compressed

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To determine the distance the spring is compressed when a 2 kg block is dropped from 0.45 meters, conservation of energy principles must be applied. The gravitational potential energy of the block converts into elastic potential energy in the spring at maximum compression. The equation used is Ui + Ki = Uf + Kf, where the initial potential energy is mgh and the final kinetic energy is zero at maximum compression. The maximum compression can be calculated using the formula .5mv^2 = .5kx^2, factoring in the gravitational force. Understanding the relationship between gravitational potential energy and spring compression is crucial for solving the problem accurately.
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A 2 kilogram block is dropped from a height of .45 meter above an uncompressed spring. The spring has an elastic constant of 200 Newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it.

A) Determine the distance that the spring is compressed
B) Determine the maximum compression of the spring
C) Determine the amplitude of the SHM

Mass = 2kg
Height = .45m
k = 200N/m

for A...I think I have to use conservation of energy, but I don't know what to put for the maxmium speed of the block.
 
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When the spring is compressed completely, how fast is the block moving?
 
robb_ said:
When the spring is compressed completely, how fast is the block moving?

the speed would be 0, right? at max. compression
 
yes, that is right
 
so...do i use

Ui + Ki = Uf + Kf
.5mv^2=.5kx^2
mv^2=kx^2...

? if i do, what's the velocity?
 
What other types of potential energy do you know of?
 
That would be true for a horizontal situation, but you have to take gravity into account for this problem. Dont be intimidated by the quadratic
 
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