Solve Force & Motion Homework: Find Stretch of Spring w/ 2 & 3 kg Masses

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To solve the problem of spring stretch with a 2.0-kg and a 3.0-kg mass connected by a spring on a frictionless surface, a 15-N force is applied to the larger mass. The total force acting on the system is 15 N, but the weights of the masses are irrelevant since they act vertically. The spring force is internal and does not affect the net force on the system. To find the acceleration, one must consider the total mass and the applied force. The correct calculation leads to a spring stretch of 4.3 cm, as stated in the book answer.
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Homework Statement


A 2.0-kg mass and a 3.0-kg mass are on a horizontal frictionless surface connected by a massless spring with spring constant k = 140 N/m. The large mass is on the right of the spring: the small mass on the left. A 15-N force is applied to the large mass. How much does the spring stretch? The book answer is 4.3 cm.


Homework Equations


F = -kx
F = ma
W = mg


The Attempt at a Solution



Total force on the system is 15 N
2 kg + kx + 3 kg = 15 N
2 (9.8) + 140 x + 3 (9.8) = 15 N
Somehow I am getting the signs wrong because I cannot come up with 4.3 cm.
 
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Cantworkit said:
Total force on the system is 15 N
Good.
2 kg + kx + 3 kg = 15 N
2 (9.8) + 140 x + 3 (9.8) = 15 N
Not sure what you're doing here:
(1) The weights of the masses act vertically, not horizontally; they aren't relevant to this problem.
(2) The spring force is internal to the system, so it's net effect on the system as a whole cancels out.

Hint: Answer these questions:
What's the acceleration of the system?
What forces act on the larger mass?
 
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