Finding the Stopping Point of a Falling Mass on a Spring

AI Thread Summary
A 0.25 kg mass attached to a spring with a spring constant of 5.4 N/m is dropped, and the goal is to find the stopping point. The correct approach involves using conservation of energy, equating gravitational potential energy at the top to spring potential energy at the lowest point. The mass's weight (mg) is 2.45 N, but the calculated stretch using F=-kx was incorrect because it did not account for the sudden drop. The final stretch of the spring is twice the stretch at equilibrium due to the dynamics of the drop. The solution confirms that energy conservation leads to the correct stopping point of 0.91 meters.
zachmgilbert
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Homework Statement


A 0.25 kg mass is attached to a spring with spring constant 5.4 N/m and let fall. To the nearest hundredth of a meter what is the point where it 'stops'?

diagram here:
http://wps.prenhall.com/wps/media/objects/1088/1114633/ch11/grav.gif


Homework Equations


y=Asin(k/m)1/2t


The Attempt at a Solution


I don't know how to find t, or I am using the wrong equation.
 
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What is going on(Force wise) when the mass reaches its lowest point?
 
I tried F=-kx using mg for force. mg is (.25)(9.8) which equals 2.45. Dividing by 5.4 gave me .45 which is wrong. The correct answer is .91
 
It is a giancoli problem so I can reset the variables. I've tried it multiples times and the F=-kx is always half of the correct answer. Where does it get multiplied by 2.
 
Your equation assumes the weight is slowly lowered to its equilibrium position. In this case, however, it is dropped suddenly. Use conservation of energy to show that its total stretch is twice the stretch in its at rest equilibrium position.
 
i have .5mv2=.5kx2 but i don't know how to find velocity.
 
zachmgilbert said:
i have .5mv2=.5kx2 but i don't know how to find velocity.
The object starts from rest with no speed, and when it reaches the bottom, it momentarily stops and also has no speed. But there is gravitational potential energy at the top, and spring potential energy at the bottom.
 
Thank You, that gave me the right answer.
 
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