Accelaration of block at its lowest position will be what ?

Question

Eshan · Answer

Dear student,

Since the springs are connected in parallel, the equivalent spring constant of the system is $k=100N/m+300N/m=400N/m$
When the block reaches the bottommost point, it will lose all its gravitational potential energy into the elastic potential energy.

Hence $mgx=\dfrac{1}{2}kx^2$
$\implies x=\dfrac{2mg}{k}=\dfrac{2\times 2\times 10}{400}m=0.1m$
At this point, the net force acting on block upwards= $kx-mg=40N-20N=20N$
Hence acceleration of the block upwards= $\dfrac{F}{m}=10m/s^2$

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Accelaration of block at its lowest position will be what ?

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Accelaration of block at its lowest position will be what ?

Accelaration of block at its lowest position will be what ?

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