Springs and friction and energy, etc.

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The discussion focuses on a physics problem involving a block compressed against a spring, which then slides through a loop. The key equations include the potential energy of the spring, kinetic energy, and gravitational potential energy. The block's energy conservation is central to solving for the compression distance, x, in terms of the spring constant k, mass m, loop radius r, and gravitational acceleration g. Participants clarify that the problem's title mistakenly includes "friction," as friction is neglected in the scenario. The conversation emphasizes the importance of correctly applying energy conservation principles to arrive at the solution.
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Homework Statement


A small block of mass m is pressed up
against a spring. The spring has constant k
and is compressed a distance x. The block
slides along a track up and around a loop
of radius r and down and out on the far
side, as shown. At the top of the loop, the
loop exerts a normal force on the block
equal to the weight of the block. Neglect
friction.

Determine the compression distance, x, in terms of given quantities k, m, r and g.

Homework Equations



U_{spring} = 1/2kx^{2}
KE = 1/2mv^{2}
U_{grav} = mgh

The Attempt at a Solution



I started with energy conservation, but I am not sure if you just set U_{spring} equal to the sum of KE and U_{grav}

The picture of the problem can be seen here: http://img137.imageshack.us/img137/7804/41935428tm6.th.jpg
 
Last edited by a moderator:
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I'm not sure if I understood the problem. Why is the problem called Springs and FRICTION and energy?

However, I would start with the loop:

You know that in the look, the gravitational potential is equal to the energy of the block. Therefore:

E_{kin}+m\cdot g\cdot h\cdot= U_{grav}

where h is equal to two times the radius r.

Now, you can set up another equation because you know that the energy of the block at the beginning was zero (no kinetic energy and no gravitational energy):

E_{kin}= 1/2\cdot k\cdot x^{2}

Now, you should know how to solve the problem...
 
mistake on title

yea sorry about the title, it wouldn't let me edit it though. :-p

thanks for the help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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