Solving Hooke's Law Problem: 2kg + 3kg Masses, k=136N/m, F=18N

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SUMMARY

The problem involves a 2kg mass and a 3kg mass connected by a massless spring with a spring constant of k=136N/m, subjected to an applied force of 18N. The initial attempt to calculate the spring stretch using F=-kd resulted in an incorrect value due to neglecting the effect of both masses. The correct approach requires considering the net force acting on the system and utilizing a free body diagram to account for the forces on both masses.

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  • Understanding of Hooke's Law (F=-kd)
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  • Basic principles of Newton's second law of motion
  • Ability to solve equations involving multiple masses
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Homework Statement


A 2kg mass and a 3kg mass are on a horizontal frictionless surface, connected by a massless spring with spring constant k=136N/m. An 18 N force is applied to the larger mass. How much does the spring stretch from its equilibirum length? (answer in cm)

Homework Equations


F=-kd
k= 136N/m
F=18N


The Attempt at a Solution


I did the obvious, I used the equation above and solved for D: 18/136 = 0.132m x 100= 13.2 cm, which is wrong. I obviously neglected the two masses, which is probably where I went wrong? Can anyone help me out by providing the proper equation and solution?
 
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Did you try drawing a free body diagram of the two mass strings connected by a massless string? It might be easier for you to write down your equations for this problem after you draw the free body diagram. Do you agree that the spring force and the applied force make up the net force?
 

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