Finding the Stopping Point of a Falling Mass on a Spring

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SUMMARY

The discussion focuses on calculating the stopping point of a 0.25 kg mass attached to a spring with a spring constant of 5.4 N/m after being dropped. The correct stopping distance is determined to be 0.91 meters, achieved by applying the conservation of energy principle. The participant initially misapplied Hooke's Law (F = -kx) and did not account for the energy transformations involved. The final solution involves equating gravitational potential energy and spring potential energy to find the correct stretch of the spring.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of gravitational potential energy (PE = mgh)
  • Familiarity with spring potential energy (PE = 0.5kx²)
  • Basic principles of conservation of energy
NEXT STEPS
  • Study the conservation of energy in mechanical systems
  • Learn about the dynamics of oscillating systems and damped motion
  • Explore advanced applications of Hooke's Law in real-world scenarios
  • Investigate the effects of mass and spring constant on oscillation frequency
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators looking for practical examples of energy conservation and spring dynamics.

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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