## Work and Conservation Laws

#### Conservation Laws

Most people remember some version of the following statements as a statement of a conservation law. The example will be for energy, but it could be mass, momentum, or anything else that is conserved.

1. The total amount of energy in the universe is constant.
2. Energy can't be created or destroyed, it can only change form.
3. The total amount of energy you start with must equal the total amount of energy you end up with.

In this section we will focus on the last statement for both energy and momentum. Energy and momentum must be conserved at the same time during any collision or other event. This is significant because one form of energy (kinetic energy) and momentum both have the same velocity in their mathematical models. But first we must deal with the concept of work.

#### Work

Work is the energy required to move an object. The farther the object is moved the more work will be expended. Work is also dependent on the force needed to make the object move. If the force is constant and in the same direction as the motion, then the work can be expressed as work equals force times displacement or W = Fs; where W is the work, F is the force, and s is the distance moved (displacement).

The amount of work (energy) needed to raise an object with mass m to a height h would be the force, which is against gravity, mg, times the height (h) or mgh. This new energy would be stored in the mass and is called the gravitational potential energy. PEgravity = mgh. This can also be thought of as finding the area under the line generated by graphing force versus distance, which in this case would be be a horizontal line and the work would correspond to the area of the square under that line.

When stretching a spring the force is not constant, but it increases linearly with the distance stretched. The amount of force needed also depends on the tightness of the spring, which is expressed by the spring constant k. The force is then given by the spring constant times the distance stretched or kx. The force is always toward the equilibrium position, so it is often expressed as -kx. To get the energy needed to stretch a spring (or the work) we find the area under the line generated in this case, which would result in a triangle with one side being the maximum force and the other side being the distance that the spring is stretched. The area under that curve is one-half of the maximum force times the distance stretched or (½)(kx)(x) = ½kx². That energy is stored in the spring and is the potential energy of the spring. PEspring = ½kx².

For situations when a frictional force, fk, is present;

W = fks = µkFns

where µk is the coefficient of kinetic friction, Fn is the normal (perpendicular) force, and s is the distance traveled, or the distance that friction exerts a force.

#### Work Example

How much work is done when a 3.5 kg box slides down a 40° friction-free incline a distance of 3 m?

The direction of the 3 m displacement is along the incline. The force along the incline, in this case, is the component of the weight of the box along the incline, given by Fx = (3.5 kg)(9.8 m/sec²)(sin40) = 22.05 N. The work, which has units of N-m or J (Joules), is W = Fs = (22.05 N)(3 m) = 66.15 J.

#### Conservation of Energy

The two categories of energy that most people are aware of are kinetic energy, KE, usually called the energy of motion, and potential energy, PE, which is stored energy due to position or structure.

When dealing with frictional forces, which produce heat, another category of energy, labeled W, is needed. W can be calculated for frictional processes by calculating the work from the frictional force and the distance that the force is applied, work equals frictional force times distance. Note that work and energy are intimately related; in fact, energy is often defined as the ability to do work. The law of conservation of energy, in classical mechanics, simply states that the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy plus W. If other forms of energy are produced (sound, light, etc.), they are incorporated into W. The mathematical model for conservation of energy is:

KEi + PEi = KEf + PEf + W

The kinetic energy depends on mass and velocity:

KE = ½mv²

where ½ is a proportionality constant, m is the mass and v is the velocity. The initial kinetic energy is obtained by using the initial velocity, vi. The final kinetic energy is obtained by using the final velocity, vf.

The gravitational potential energy, which will be used here for potential energy, depends on the mass and the height:

PE = mgh

where m is the mass, g is the acceleration due to gravity (9.8 m/sec²) and h is the height. The initial potential energy is obtained by using the initial height, hi, and the final potential energy is obtained by using the final height, hf. The position where h equals zero is an arbitrary choice. Since there is on absolute zero for height the change in position and the change in potential energy are the only measurable quantities.

When doing these problems remember that when a frictional force, fk, is present;

W = fks = µkFns

where µk is the coefficient of kinetic friction, Fn is the normal (perpendicular) force, and s is the distance traveled, or the distance that friction exerts a force.

#### Conservation of Energy Example

A 20 kg box starts from rest and slides 5 m down a 30° friction-free incline. (a) What is the final kinetic energy and the final velocity of the box? (b) What will the kinetic energy and velocity be if µk = 0.10?

KEi + PEi = KEf + PEf + W

½mvi² + mghi = ½mvf² + mghf + W

In part (a) the initial velocity is zero (starts from rest), the initial height is (5 m)(sin30), the final height is zero, and the work is zero (no friction). Also, the mass is 20 kg and g = 9.8 m/sec².

0 + (20 kg)(9.8 m/sec²)(5 m)(sin30) = ½(20 kg)vf² + 0 + 0

When there isn't any friction, the initial PE is equal to the final KE. Notice that the mass cancels out, giving

vf = [(2)(9.8 m/sec²)(5 m)(sin30)]½ = 7 m/sec and

KEf = ½mvf² = ½(20 kg)(7 m/sec)² = 490 J

For part (b) there is friction which means that W is not zero. For a ramp, where Fn is the y-component of the gravitational force Fg = mg,

W = fks = µkFns = µk[(mg)cos30)]s

Plugging in the numbers:

0 + (20 kg)(9.8 m/sec²)(5 m)(sin30) = ½(20 kg)vf² + 0 + (0.10)(20 kg)(9.8 m/sec²)(cos30)(5 m)

The mass again cancels and

vf = {2[(9.8 m/sec²)(5 m)(sin30) - (0.10)(9.8 m/sec²)(cos30)(5 m)]}½ = 6.36 m/sec and

KEf = ½mvf² = ½(20 kg)(6.36 m/sec)² = 405 J

#### Conservation of Momentum

Momentum, p, is a measure of the tendency of an object to remain in motion, if in motion, or to remain at rest, if at rest. Momentum is a vector that is directly proportional to both mass (its harder to move a large mass) and velocity (its harder to deflect an object that is moving fast);

p = mv

Momentum can change due to a variety of sources including friction and accelerations due to collisions. The total momentum before a change must always equal the total momentum after the change. This is one statement of conservation of momentum. For a system in 1-D with two masses, m1 and m2, that goes from an initial state, i, to a final state, f, the mathematical model is

p1i + p2i = p1f + p2f

m1v1i + m2v2i = m1v1f + m2v2f

In two dimensions the momentums and velocities would be vectors. The momentum vector can be resolved in x- and y-components in the usual way, and the conservation laws can be applied separately to each of the perpendicular directions.

If there is a loss of energy during a collision due to friction or other causes, making W nonzero, the collision is inelastic. When there is no such loss of energy, making W zero, the collision is elastic.

#### Momentum Examples

A 45 kg ice skater is moving at 2 m/sec while a second 60 kg skater is coming directly toward her at 3 m/sec. Upon collision they hug each other as if they were old friends and remain together as they move. What will their final velocity be?

m1v1i + m2v2i = (m1+m2)vf

vf = (m1v1i + m2v2i)/(m1+m2)

vf = [(45 kg)(2 m/sec) + (60 kg)(3 m/sec)]/(45 kg + 60 kg) = 2.57 m/sec

Now assume that the two skaters were coming at each other at 90o angles (one traveling North and one traveling East). What is the new magitude and direction of their final velocity in this case?

#### Conservation of Energy and Momentum

Conservation of energy and conservation of momentum are useful independently, but many problems require the use of both conservation laws simultaneously. If only initial velocities and momentums are known, conservation of momentum, used alone, would allow an infinite number of final momentum pairs (the total momentum could be divided an infinite number of ways) which would also mean an infinite number of velocity pairs. Only one solution, however, will also show conservation of energy to be satisfied. The two "laws" must be used together to provide two equations with two unknowns.

#### Energy and Momentum Example

A 500 g billiard ball going 0.3 m/sec collides head-on and elastically with a stationary ball that has twice the mass. What are the final velocities of each ball?

From conservation of energy when W = 0 (elastic collision), initial and final height is at zero, m2 = 2m1, v2i = 0:

½m1v1i² + ½m2v2i² + m1gh1i + m2gh2i = ½m1v1f² + ½m2v2f² + m1gh1f + m2gh2i + W

½m1v1i² + 0 + 0 + 0 = ½m1v1f² + ½(2m1)v2f² + 0 + 0 + 0

v1i² = v1f² + 2v2f²

From conservation of momentum:

m1v1i + m2v2i = m1v1f + m2v2f

m1v1i + 0 = m1v1f + 2m1v2f

v1i = v1f + 2v2f

Putting in the initial velocity gives two equations and two unknowns:

(0.3)² = v1f² + 2v2f²

0.3 = v1f + 2v2f

Solve for v1f in the second equation, substitute into the first equation, and then solve for v2f:

v1f = 0.3 - 2v2f

(0.3)² = (0.3 - 2v2f)² + 2v2f²

(0.3)² = (0.3)² - (2)(0.3)(2v2f) + (2v2f)² + 2v2f²

0 = -(2)(0.3)(2v2f) + (2v2f)² + 2v2f² = [-(2)(0.3)(2) + 4v2f]v2f

0 = [-(1.2) + 4v2f]v2f

The two roots of the equation are v2f = 0, or v2f = 0.30. It is physically impossible for v2f to be zero. There is no way for conservation of momentum to be satisfied in that case. So,

v2f = 0.30 m/sec and v1f = 0.3 - 2v2f = 0.3 - 2(0.3) = -0.3 m/sec

This satisfies conservation of momentum since the initial momentum is (0.5 kg)(0.3 m/sec) = 0.06 kg-m/sec and the final momentum is (0.5 kg)(-0.3 m/sec) + (1.0 kg)(0.3 m/sec) = -0.06 + 0.3 = 0.06 kg-m/sec. The first ball hits the second ball, bounces directly backward at the same time causing the second ball to move forward. The total momentum before is equal to the total momentum afterward. You can also verify that conservation of energy is satisfied.