Finding the Stopping Point of a Falling Mass on a Spring

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Homework Help Overview

The problem involves a 0.25 kg mass attached to a spring with a spring constant of 5.4 N/m, which is let fall to determine the stopping point of the mass when it compresses the spring. The context includes concepts from mechanics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass at its lowest point and the application of Hooke's Law. There are attempts to use energy conservation principles, with some participants questioning the assumptions made about the system's behavior when the mass is dropped.

Discussion Status

Some participants have provided insights into the relationship between gravitational potential energy and spring potential energy, suggesting that the total stretch of the spring is greater than the equilibrium position due to the initial drop. There is a recognition of differing interpretations regarding the application of formulas and the conditions of the problem.

Contextual Notes

Participants note the challenge of determining velocity at different points in the motion and the implications of starting from rest versus being dropped. There is mention of the problem being from a specific textbook, which may influence the approach taken.

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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