Conservation of Energy Example 3

Guide 8-6c. Conservation of Energy Example 3

Here is an example of a problem requiring the solution of a quadratic equation. A block of mass 0.50 kg is dropped from rest from a height of 0.75 m above a spring of spring constant 25 N/m. What is the greatest distance that the block compresses the spring?

Given:

We need 3 different heights. y₁ is the initial height of the block; y₂ is the equilibrium position of the spring, and y₃ is at the maximum compression of the spring. Note that we select the equilibrium position to be the zero of position. We also know:

v₁ = 0
y₁ = 0.75 m
v₃ = 0
m = 0.50 kg
k = 25 N/m

Goal: Find |y₃|. Note that we're looking for the absolute value of y₃. The latter will be a negative number, since y₃ is below the point selected for the origin.

system = block, Earth, spring

external forces: none

initial state = block at height y₁
final state = spring completely compressed

Note that it won't be necessary to find the speed of the block at y₂ = 0. The fact that both the initial and final kinetic energies will be zero will simplify the problem.

ΔU_g is negative since the block decreases in elevation;
ΔU_e is positive since the spring compresses;
ΔK is 0 since the initial and final velocities are 0.

The initial and final kinetic energies are both 0. There are 2 potential energy changes. The system loses gravitational potential and gains elastic potential energy.

We get a quadratic equation in the unknown y₃. Solving the equation yields two roots. We select the negative root, because we know y₃ is negative based on the way the y-axis was set up. The greatest distance of compression is 0.77 m.

One check is to see what the equation reduces to when y₁ = 0. That is, the block is released at the top of the spring. In that case, the lost gravitational potential energy mgy₃ goes into spring energy ½ky₃². Solving for y₃ gives -2mg/k, as the formula to the left predicts with y₁ = 0.