a 2 kg mass and a 3kg 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. How much does the spring stretch from its equilibrium length? The masses uniformly accelerate with no oscillations.

I got the answer is change in x = 0.429 cm, am I right?

I get the same answer. Well done :)

I get a slightly different answer. I calculate the tension in the spring to be 6N, so its extension is 6N/(140N/m) = 0.0429m, or 4.29cm.

Quote:

---

Originally Posted by ebaines

I get a slightly different answer. I calculate the tension in the spring to be 6N, so its extension is 6N/(140N/m) = 0.0429m, or 4.29cm.

---

You're right, I think I misread my power of 10 in my calculator :(