Extension in Spring Homework: Find x

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The discussion revolves around calculating the extension of a spring using conservation of energy and momentum principles. The initial calculations yield a spring constant (k) of 200 and a velocity of the block at 3 m/s, leading to an extension (x) of 4.5 cm. However, the expected answer is 6.1 cm, indicating that gravitational potential energy (GPE) must also be considered as both masses lose GPE when the spring extends from equilibrium. The equation provided highlights the need to account for the GPE in the energy balance. Accurate calculations require integrating both kinetic and potential energies to determine the correct extension.
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Homework Statement


attachment.php?attachmentid=32924&stc=1&d=1299767447.jpg



The Attempt at a Solution



first using the initial extension i found out that k = 200.

velocity of block = √(2gh) = √(2*10*.45) = 3

then using conservation of momentum ...

120 * 3 = 320 * V
V = 1.125

this kinetic energy converts into potential energy of spring

.5 * .32 * V = .5 * 200 * x
x = 4.5 cm

but answer is given as 6.1 cm
 

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You haven't taken into account the gravitational potential energy, GPE is lost by both masses as the spring extents from equilibrium.


<br /> <br /> \frac{1}{2} m v^2=\frac{1}{2}kx^2 -mgx<br /> <br />
 
Thanks for your help ... :)
 
Thread 'Variable mass system : water sprayed into a moving container'

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