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Force and motion

  • Thread starter Cantworkit
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  • #1
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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.
 

Answers and Replies

  • #2
Doc Al
Mentor
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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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