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    western50's Avatar
    western50 Posts: 105, Reputation: 1
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    #1

    Feb 5, 2011, 11:11 PM
    force by spring
    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?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Feb 6, 2011, 03:59 AM

    I get the same answer. Well done :)
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Feb 7, 2011, 08:13 AM

    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.
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    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Feb 7, 2011, 08:50 AM
    Quote Originally Posted by ebaines View Post
    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 :(
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