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A 10-kg block is attached to a vertical spring of constant 2000 N/m and slowly lowered...

A 10-kg block is attached to a vertical spring of constant 2000 N/m and slowly lowered to its equilibrium position. The block is now pulled down a distance of 3 cm and released from rest and executes vertical SHM. What is the net force on the block when it is at its lowest position? (Assume g = 10 m/s2.)

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Answer #1

At the lowest position only potential energy exists and kinetic energy is zero.

Since we have given the elongation from the equilibrium position itself(after the mass being attached), we need not to consider the effect of graity to solve this problem.

So, Net force on the block at the lowest position = Maximum Potential energy = 1/2 × K × x2

K- Spring constant = 2000N/m

x - elongation = 3 cm = 0.03 m

F = 1/2 × 2000 × (0.03)2 = 0.9 N

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