Potential energy. What is the spring constant?

1. Feb 24, 2015

Y*_max

1. The problem statement, all variables and given/known data
Figure 8-36 shows an 8.00 kg stone at rest on a spring. The spring is compressed 10.0 cm by the stone. (a) What is the spring constant?

2. Relevant formula
Mechanical energy is conserved

3. The attempt at a solution
The decrease in gravitational potential energy that occurs when the block is put on the spring (the spring is compressed) is equal to the increase in elastic potential energy of the spring.
Thus: mgh=0.5kh^2 (the coordinate system is chosen so that gravitational potential energy is zero when the block is at rest and the spring is compressed)
Solving the equation for k gives k=2gm/h
k=2*9.8*8.0/0.1=1568 N/m.

Yet, the answer turns out to be 784N/m :/
What did I do wrong?
Thanks!

2. Feb 24, 2015

AbhinavJ

The downward force on the force is equal to mg.
The restoring force by the spring in opposite direction is kx, equating
8g=k/10
K=9.8*8*10=784N

3. Feb 24, 2015

Suraj M

its 10 cm so you should take 0.1m , And you missed a factor of 2

not necessarily true! There is no diagram! Anyway, try equating mg = kx , you won't get the same answer!! Was the body dropped suddenly or slowly lowered down to equilibrium?

Last edited: Feb 24, 2015
4. Feb 24, 2015

BvU

Max' question "what did I do wrong ?" still stands !

You did an energy balance for a situation where all potential energy from gravity is converted into mechanical energy to compress the spring.
That would be: let go of the stone at the top of the uncompressed spring and see where the mass stops moving (i.e. the motion reverses direction). However, at that point the energy in the spring is enough to push the stone back to the original postition (again with potential energy mgh and no kinetic energy) where the sequence would repeat.

That is not what was given in the exercise. The stone was lowered gently until equilibrium position. The hand that lowered the stone took away half the potential energy.

Last edited: Feb 24, 2015
5. Feb 24, 2015

BvU

No, it definitely is not. Max did the $k=\frac{2mgh}{h^2}$ just fine, only the energy balance didn't apply for the situation described.

6. Feb 24, 2015

Suraj M

I realised that 12 mins ago,BvU changed it, sorry!

7. Feb 26, 2015

Y*_max

Oh, I see! Thank you very much to you all!