Energy Conservation Spring Compression Problem

[redcard]

This is Problem 10.46 from the Engineering Physics Book by Knight.

A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it.

a.) How far does the spring compress if you let go of the block suddenly?

b.) How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?

I have missed class 3 days in a row (sick) and do not know what to do at all. That being said, thorough answers are appreciated.

 

[rl.bhat]

Fall of potential energy of the block=energy stored in the spring. Assume block moves x m. The block moves the same distance whether you release the block suddenly or slowly.

 

[Shooting Star]

a.) You have to equate the initial and final energies.
b.) You have to equate the weight of the body to the upward force by the spring. In the 2nd case, you are not allowing the mass to transfer all its PE to KE.

 

[Shooting Star]

Fall of potential energy of the block=energy stored in the spring. Assume block moves x m. The block moves the same distance whether you release the block suddenly or slowly.

That is not correct.

 

[FedEx]

Here you must consider that the potential energy of the block gets converted to the elastic potential energy of the spring. Which let's say gets compressed by x meters.

Hence

$$\frac{kx^2}{2} = mgx$$

Here in case of the right hand equation we will consider only x as the block is left from almost negligible height.

 

[Shooting Star]

Here you must consider that the potential energy of the block gets converted to the elastic potential energy of the spring. Which let's say gets compressed by x meters.

Hence

$$\frac{kx^2}{2} = mgx$$

Here in case of the right hand equation we will consider only x as the block is left from almost negligible height.

Does this cover both the cases, according to you? What about the fact that I am lowering the mass in the 2nd case by holding it in my hand?

 

[rl.bhat]

Spring problem

When I said fall of potential, I ment that the block is moved in the down ward direction.
In the 2nd part you are removing the hand when the block remains indisturbed. And it is the position of the maximum compression of the spring. And it is the same wherher you remove the hand suddenly or slowly.

 

[Shooting Star]

Please read my earlier post. The answers to the two questions asked by the OP are different.

In case b, the block on the spring will stay at some point, when equilibrium is achieved.

In case a, the spring is compressed beyond that point and then again bounces up. Without friction or dissipation of energy, it will continue to oscillate, with the position in case a as the mean position.

Let the OP give us the final values.