Finding the speed of a dropped block using energy conservation

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SUMMARY

The discussion revolves around calculating the speed of a 2.40 kg block dropped from a height of 5.00 m onto a spring, which compresses by 25.0 cm when the block is momentarily at rest. The key equations involved are the potential energy (PE = mgh), kinetic energy (KE = 1/2mv^2), and spring energy (E = 1/2kx^2). To find the speed of the block when the spring is compressed by 15.0 cm, users must first determine the spring constant (k) using the energy conservation principle, equating the total potential energy lost to the energy stored in the spring.

PREREQUISITES
  • Understanding of energy conservation principles in physics
  • Familiarity with Hooke's Law (F = -kx)
  • Knowledge of potential energy (PE = mgh) and kinetic energy (KE = 1/2mv^2)
  • Ability to manipulate algebraic equations to solve for unknowns
NEXT STEPS
  • Calculate the spring constant (k) using the energy conservation method
  • Learn how to derive the speed of an object using energy equations
  • Explore examples of blocks falling onto springs in physics textbooks
  • Study the relationship between compression distance and energy stored in springs
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of practical applications of these concepts.

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Homework Statement


A 2.40 kg block is dropped onto a spring and platform of negligible mass. The block is released a distance of 5.00 m above the platform. When the block is momentarily at rest, the spring is compressed by 25.0 cm. Find the speed of the block when the compression of the spring is only 15.0 cm.

I need some help setting this up...

Homework Equations



F= -kx

PE = mgh

KE = 1/2mv^2

The Attempt at a Solution



I started off with mgh = -kx + 1/2mv^2 but I don't know what the value of k is so I can't solve for v (I don't have the answer, but I don't think the letter k would be part of it). I know the distance between where the block starts and compression at 15.0cm is 5.15cm so I think I use that for x? really I just have no idea how to do this and I can't find an example in my book of a block falling on a spring. help!
 
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You must also consider the energy stored in the spring
<br /> E = \frac{1}{2}kx^2<br />
The total potential energy lost by the box during the fall, is equivalent to the energy stored in the spring at the moment when the box is at rest. This will allow you to calculate k.

Then use the total energy as
<br /> E_T=mgh+\frac{1}{2}mv^2+\frac{1}{2}kx^2<br />
to find the speed of the block when x = 15cm
 
jensson said:
I started off with mgh = -kx + 1/2mv^2 but I don't know what the value of k is so I can't solve for v (I don't have the answer, but I don't think the letter k would be part of it). I know the distance between where the block starts and compression at 15.0cm is 5.15cm so I think I use that for x? really I just have no idea how to do this and I can't find an example in my book of a block falling on a spring. help!

Firstly in you equation mgh = -kx + 1/2mv^2 you are equating force with energy. It should be 1/2 kx2.

Also, you can calculate k from the given data. Just try it!
 

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