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srinidhi321 · Aug 22, 2012, 05:35 AM

Two Blocks of mass M1 and M2 are connected by a spring of force constant k?
If the block 1 is elongated towards right to a distance X1 and Block 2
is elongated towards left to a distance of X2 simultaneously . Then
what is the work done by the spring on each of these blocks
"separately"?

ebaines · Aug 22, 2012, 06:31 AM

I'm going to assume that you are familiar with calculus - if not, please post back and we'll do this a different way:

The work done on the spring is the integral of the force applied by the spring with respect to the distance the force operates over:

$ W = \int F dx $

For this problem the force applied by the spring is the spring constant times its total extension. If we let $x_R$ be the extension to the right and $x_L$ the extension to the left, then the total extension e is $e = x_R + x_L$ . Now we need to find the total extension as a function of the position of the right hand block. Since we know that both blocks move simultaneoulsy at a steady rate, and we know that the total extension is $e_{final} = X_1 + X_2$ , then we can say that e as a function of $x_R$ is equal to:

$e = x_R+ x_L = x_R (1 + \frac {X_2}{X_1})$ .

The work integral for the right hand block is:

$ W_1 = \int F dx_R = \int_0 ^ {X_1} kedx_R = \int_0 ^{X_1} kx_R (1+ X_2/X_1)dx_R= \frac 1 2 k X_1^2(1+X_2/X_1) = \frac 1 2 k X_1^2 + \frac 1 2 k X_1X_2 $

Now apply the same method for the block on the left. As a check that you've done it properly you should find that if you add the two components of work it is equivalent to the work done if just one block is moved a distance $X_R+X_L$ while the other is held stationary.