Guide to Solving Conservation of Energy Problems, Part III

Additional problem types

We'll complete this guide to solving conservation of energy problems with some problem types that bring in additional elements, such as connected objects and circular motion, or additional algebraic complexity, such as situations leading to quadratic equations.

When more than one object is involved such as in a connected object problem, you simply include kinetic and potential energy contributions from all the objects.  See Example 8-10 in the text for a connected object problem.

Below is an example of a combined energy-circular motion problem.  The method of solution shown can also be applied to pendulums and swings where you're asked to find the tension in a string or rope.  The problem is number 8-52 on page 236 (2nd ed: 8-50 on page 229). We'll give complete solutions but provide fewer comments in blue than for previous examples.

Given:

vA = 8.0 m/s
yA = 1.75 m
yB = 0
m = 61 kg
r = 12 m

Goal:  Find the normal force on the skier at the bottom of the depression.

system = skier, Earth, gravity

external forces: normal

initial state = skier at point A
final state = skier at point B
First, use conservation of energy to solve for the speed of the skier at the bottom of the depression.
Next, do a net force problem to find the normal force.  The acceleration is up (toward the center of the circle) at the bottom of the depression, so the direction of positive is selected pointing up.

The net force equation is solved for N.  Then the result for vB² from the previous conservation of energy problem is substituted.

The result has units of newtons and is positive as expected for a force magnitude. The value of 1100 N (about 250 lbs) seems high. This is greater than the weight of the skier. But then, the depression is fairly deep.

Next 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, gravity

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.

 
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.