The 10 kg block is attached to a spring k 20 Ncm In state 1

The 10 kg block is attached to a spring (k = 20 N/cm). In state 1, the spring is stretched 10 cm and released from rest. What is the distance d if the block has zero velocity in state 2?

Solution

>> Let\'s first find the stretch in spring at equilibrium point.

Let it be \"x\"

So, under Equilibrium,

mg = kx

Putting Values,

m = mass = 10 kg

k = Spring Constant = 20 N/cm = 2000N/m

Putting all above,

10*9.81 = 2000*x

=> x = 0.04905 m = 4.905 cm

and as it is totally stretched by 10 cm

>> Now, considering this system as a Simple Harmonic Motion,

y = Max distance from Equilibrium at which oscillation starts.

Here, y = 10 - x = 5.095 cm

So, at top when block has zero velocity again, it will be now at top extreme point and at \"y \" distance from Equilibrium point.

So, d = 2y = 10.19 cm. ...... ANSWER.....