Two blocks, with masses M = 241g and 2 M = 482g, are connected to a spring of sp

Question

Two blocks, with masses M = 241g and 2M = 482g, are connected to a spring of spring constant k = 16.2N/m that has one end fixed, as shown in the figure. The horizontal surface and the pulley are frictionless and the pulley has negligible mass.The blocks are released from rest with the spring relaxed.

What is the speed of the blocks when the hanging block has fallen 26.7cm? Note that this also stretches the spring by 26.7cm.

How far will the hanging mass fall before momentarily coming to rest?

What is the maximum speed of the blocks?

Explanation / Answer

one block falls by 26.7cm and other block moves due to which the spring streches.

So net change in Potential Energy = +(0.5)*(16.2)*(0.267)^2 - 0.482*9.8*0.267

Change in K.E = 3*(0.5)*0.241*v^2

Since energy is conserved, -Chage in P.E = change in K.E

So v^2 = 1.892

v = 1.375m/s

If the block comes to a momentarily stop then Change in K.E =0

So 0.5*16.2*(x)^2 -0.482*9.8*x = 0

Solving gives x = 0 or 0.582

So spring extends by 5.82cm

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