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Since the block is motionless, we know that our forces will cancel out:

Fnet=0

There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:

$F_{\text {spring,top}}+F_{\text {spring}, b o t}-m g=0$
Plugging in expressions for each spring force, we get:
$k x_{\text {top}}+k x_{\text {bot}}-m g=0$
Rearring for the displacement of the top spring, we get:
x_{t o p}=\frac{m g-k x_{b o t}}{k}=\frac{(3 k g)\left(10 \frac{m}{s^{2}}\right)-\left(25 \frac{N}{m}\right)(0.4 m)}{25 \frac{N}{m}}$\\ $x_{t o p}=\frac{30 N-10 N}{25 \frac{N}{m}}=0.8 m$

Posted by

Satyajeet Kumar

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