What Is the Mechanical Energy of a Spring-Mass System?

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SUMMARY

The mechanical energy of a spring-mass system can be calculated using the spring constant and the displacement from equilibrium. In this case, a 0.766-kg mass attached to a spring with a spring constant (k) of 78.0 N/m was stretched 25.53 cm, resulting in a total mechanical energy of 2.54 J. The maximum speed of the mass was incorrectly calculated as 3.64 m/s, while the correct maximum speed is 2.58 m/s. The work-energy theorem and the equations of motion were essential in determining the speed at various displacements.

PREREQUISITES
  • Understanding of Hooke's Law (Fs = -kx)
  • Familiarity with kinetic energy formula (K = (1/2)mv^2)
  • Knowledge of the work-energy theorem
  • Basic calculus for integrating work done by a spring
NEXT STEPS
  • Review the work-energy theorem in detail
  • Learn about energy conservation in spring-mass systems
  • Study the integration of work done by a spring
  • Practice solving problems involving oscillatory motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy systems, as well as educators looking for problem-solving strategies in spring-mass dynamics.

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Homework Statement


A 0.766-kg mass is attached to a horizontal spring with k = 78.0 N/m. The Mass slides across a frictionless surface. The spring is stretched 25.53 cm from equilibrium, and then the mass is released from rest.
(a) Find the mechanical energy of the system.
(b) Find the speed of the mass when it has moved 3.63 cm.
(c) Find the maximum speed of the mass.


Homework Equations


Fs=-kx
K=(1/2)mv^2
Work = Integral( Kx) dx from a to b



The Attempt at a Solution


F=-(78.0 N/m * .2553 m) = 19.9134 N
W=(78x^2)/2 from 0 to .2553 = 2.54 J

I'm not sure where to go from here..
 
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After I posted this a light went on and I used F=ma to find a = 25.99 m/s. I then used Vf^2 = Vi^2 + 2ad to find the answer to b and c, however I still got part c wrong for some reason. for part b I got Vf = sqrt(2(25.99)(.0363))=1.37. For part c I got Vf = sqrt(2(25.99)(.2553))=3.64 m/s/s. According to the webwork the correct answer to part c is 2.58. Does anyone know what I am doing wrong? Thanks!
 
Work done by a spring is W=-(1/2)*k*x^2 from x_initial to x_final. Work done is also equal to the change in kinetic energy (by the work-energy theorem). These two pieces of information should enable you to find the velocity of the mass after it has moved 3.63 cm.

I hope this helps.
 

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