Spring and Block Mechanical Energy

## Question:

I don't understand how to solve the following problem:
A 2.4kg block is dropped from a height of 5.0 m onto a spring of spring constant 3955N/m. When the block is momentarily at rest, the spring has compressed by 25cm. Find the speed of the block when the compression of the spring is 15.0cm.

Thanks!

## Answer:

Neglecting all friction, the mechanical energy of the system is conserved. The mechanical energy is the sum of the potential and kinetic energy. In the initial conditions the block is at rest at a height of 5 meters above the spring. Since the block is at rest all the mechanical energy is potential energy, pe=m*g*h = 2.4*9.8*5 = 117.6 Joules. As the block falls this potential energy gets converted to kinetic energy so that the kinetic energy at the time the block strikes the spring is 117.6 Joules. As the spring slows the block, the kinetic energy is converted back into potential energy stored in the compressed spring until at a compression of 0.25 meters the block again has no kinetic energy.
The question asks what is the velocity of the block at a 0.15 meter spring compression. If we could find how much energy is required to compress the spring 0.15 meters we would be on our way, since the difference between the total 117.6 Joules and that energy must appear as kinetic energy and knowing the kinetic energy gives us the velocity.

The energy stored in a spring is the work done in compressing it. The work is force times distance as long as the force is constant. The force can only be considered constant over a very short compression, call it dx. Then the little bit of work, dw, done over that distance is force times dx, but force is the spring constant k times the total displacement x so the work in compressing a spring distance dx about the displacement x is dw=k*x*dx. The total work w in compressing a spring to any displacement x then is the sum of all the little dws. In other words w=k*Sx*dx. The way we sum up x*dx over all x is to integrate the function x*dx. To integrate, increase the exponent on x by 1 and divide by the new exponent so w=k*x^2/2 or w=1/2*k*x^2.

If x=0.15 meters then the potential energy stored in the spring is 1/2*3955*0.15^2 = 44.49 Joules. That leaves 117.6-44.49.Joules of Kinetic energy. I get 73.1 Joules. So 1/2*2.4^v^2=73.1, or v^2=60.9 m^2/s^2. That gives me a v=7.81m/s.

I should caution you at this point that my arithmetic is terrible. Never accept a numerical answer from me without doing the calculations yourself to check on me. The concepts though I believe are sound.

If you would prefer to work this without resorting to calculus, notice that the force increases linearly with displacement, as in f=k*x. This means that the average force over any distance is the simple numerical average of the initial and final forces. At a zero displacement the force is zero. At a 0.15 meter displacement the force is k*0.15. The average force is 1/2*k*.15 N. The work done in compressing the spring can be found by multiplying ths average force times the 0.15 m displacement. Again w=1/2*k*0.15^2.