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

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A problem involving a 2kg and a 3kg mass connected by a spring with a spring constant of 136N/m is discussed, where an 18N force is applied to the larger mass. The initial attempt to calculate the spring stretch resulted in an incorrect answer, as the contributor failed to account for the combined effect of both masses. Suggestions include using a free body diagram to visualize the forces acting on the system, emphasizing that the spring force and applied force contribute to the net force. The correct approach requires integrating the masses into the calculations to determine the accurate spring stretch. Understanding the dynamics of the system is crucial for solving the problem correctly.
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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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