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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. A 15-N force is applied to the larger mass, as shown in Fig. 4.23. How much does the spring stretch from its equilibrium length?

Verified Answer :

The solution would be like this
for this specific problem:

=> F/2 = (15 - F)/3 
=> 3F
= 30 - 2F 
=> 5F
= 30 
=> F =
6 N 

So if the spring
stretches by x, 
F = kx 

=> x =
F/k = 6 / 140 m = 600 / 140 cm = 4.3 cm

The spring stretches
for 4.3 cm.

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