Conservation of EnergyConservation of Energy

Potential Energy

In the previous lecture we looked at changes to a system’s energy due to an external forces. Now we’ll take a look at energies as a result of internal forces. This will lead us to a concept of an inherent energy of a body, its potential energy.

Imagine we move a book from an initial y position ya to a final position yb, as shown in the following diagram. We can assume that we do this so slowly that the kinetic energy of the book is approximately zero.

If the book is released from the higher point it will fall and gain kinetic energy. This implies that work was done on the book in raising it and furthermore that this energy is stored in the book waiting to be released. Although we gained no kinetic energy by raising the book we have stored energy in such a way as to have the potential to gain kinetic energy upon future release. We can find this energy by calculating the work done by gravity in the process.

W = (F) (Δr) cos θ = (mg)(yb – ya) (cos 1800) = mgya - mgyb

The result is an energy term which looks like an initial value minus a final value. We will call this value as the potential energy of the system.

Defining Ug as the gravitational potential energy of the system, we have

Ug = mgy

W = - ΔUg

The change in gravitational potential energy depends only on the starting and ending y coordinates, not the total distance traveled. If instead of a straight vertical distance we displaced the book in both x and y, our work would be the dot product between the force and displacement vectors:

W = F • Δr = (mgy)( j) • [(xb – xa) (i) + (yb – yb) (j)]

= (mgy) (yb – ya) = mgya - mgyb

Gravitational potential energy is dependent on the choice of origin of the y axis. The difference of potential energy (the quantity we found) is independent of the origin, and as such any convenient origin can be chosen for a problem. The origin is usually defined as the lowest vertical point in a problem for convenience.

Using our book example, let us now release the book from yb and let it fall freely back to its initial position ya. The work done by gravity in during this motion will be given by

W = (F)(Δr) (cos θ) = (mg) (yb – ya) cos 0

= mgyb – mgya

We are now going to gain kinetic energy. This increase will go according to the work-energy theorem

Wtotal = W = ΔK = mgyb – mgya = Ug(initial) – Ug(final) = - ΔUg

or

ΔK + ΔUg = 0

What we have on the left side of the equal sign is a sum of the energy stored in the system which equals zero because the system is isolated (cannot gain or lose energy due to an external transfer).

Expanding the equation we can arrive at an important conservation principle. If we expand the equation in terms of the initial and final values, as well as dropping the g subscript from U, we have

ΔK = Kf – Ki = - ΔU = Ui – Uf

Kf + Uf = Ki + Ui

We again have a set of terms for the initial and final conditions. If we define K + U = E , where E is the total mechanical energy of the system, we have

Ef = Ei

We can now state the principle of conservation of mechanical energy of an isolated system:

The total mechanical energy of an isolated system is constant*.

We insert the asterisk because this equation only holds true in the presence of what are called conservative forces.

Conservative Forces

Conservative forces allow for the conservation of mechanical energy. A conservative force has two equivalent properties:

1. The work done by a conservative force in moving between two points is independent of the path.

2. The work done by a conservative force in moving through any closed path is zero.

We can prove that gravity is a conservative force using the following figure:

We can calculate the work done in moving a 1 kg mass from point O to point C using three different paths. Assuming we are oriented such that Fg = -mg j,

Path OAC

W = (mg)(ΔrO-A) (cos 900) + (mg)(Δr)A-C (cos 1800)

= -(1 kg)(9.8 m/s2)(5m) = - 49 J

Path OC

W = (mg)(Δr) (cos (45+90)0) = (1 kg)(9.8 m/s2)(7.1 m) (cos 1350)

= - 49 J

Path OBC

W = (mg)(ΔrO-B) (cos 1800) + (mg)(Δr)B-C (cos 900)

= -(1 kg)(9.8 m/s2)(5m) = - 49 J

The work done in moving between O and C is path-independent, thus we have a conservative force by condition 1.

Another property of a conservative force is that the work done in moving between two points can be equated to the negative of the potential energy difference between the points:

W = -ΔU

We have already seen this property for gravity. Another conservative force is the spring force. Recall that the work done in displacing a spring from an initial displacement xi to a final displacement xf was given by

W = ½ kxi2 – ½ kxf

2

If we define the spring elastic energy as

Us = ½ kx2

We have

W = ½ kxi2 – ½ kxf

2 = - ΔUs

And we conclude that the spring force is conservative as well.

We can thus state that in the absence of non conservative forces the total mechanical energy, given by the sum of all potential and kinetic energy terms,

E = K + U = ½ mv2 + mgy + ½ kx2

will remain a constant:

Ei = ½ mvi2 + mgyi + ½ kxi

2 = Ef = ½ mvf2 + mgyf + ½ kxf

2

This is a more exact statement of the principle of conservation of energy for conservative forces.

A ball is dropped from rest from a height h above the ground, as shown in the figure. Air resistance can be neglected. (a) Find the speed of the ball when it reaches a vertical position of y above the ground.

Using the ground as the y = 0 position, we being to write our energy expressions for the initial state and the final state. Since only conservative forces exist, energy is conserved.

Ei = ½ mvi2 + mgyi + ½ kxi

2 = mgyi = mgh

Ef = ½ mvf2 + mgyf + ½ kxf

2 = mgy + ½ mvf2

Since energy is conserved, Ei = Ef,

mgh = ½ mvf2 + mgy

vf = (2g (h-y) )1/2

(b) Repeat the calculation if the bass has an initial upward velocity of vi.

We need to add this kinetic energy to the initial expression:

Ei = ½ mvi2 + mgyi + ½ kxi

2 = ½ mvi2 + mgh

Ef = ½ mvf2 + mgyf + ½ kxf

2 = mgy + ½ mvf2

Since energy is conserved, Ei = Ef,

½ mvi2 + mgh = ½ mvf

2 + mgy

vf = (vi2 +2g (h-y) )1/2

A simple pendulum of length L and mass m is released from rest at point A at an angle of θA with respect to the vertical. Find the speed of the mass when it reaches the bottom of the path at point B.

Using point A as our initial point and point B as our final point, and setting y = 0 at point B,

Ei = mgyi = mg (L – L cos θA) = Ef = ½ mvf2,

vf = ( 2g ( L – L cos θA))1/2

The launching mechanism of a toy gun consists of a spring of an unknown spring constant. When the spring is compressed 0.120 m, the gun, when fired vertically, is able to launch a 35.0 g projectile to a maximum height of 20 m above the position of the projectile before firing. Find the spring constant of the spring.

When the projectile is launched, it has ho kinetic energy initially, and if we set that point as y = 0, it has no gravitational potential energy. When is reaches the highest point, the spring is relaxed and the projectile has no velocity. Thus, vi, vf, yi, and xf are all zero.

Remembering that E = ½ mv2 + mgy + ½ kx2,

Ei = ½ kxi2 = Ef = mgyf

k = 2 mgyf/xi2

Using m = 0.035 kg, yf = 20m and xi = 0.12 m,

k = 953 N/m

Find the velocity of the projectile as it passes through the equilibrium position of the spring.

At the equilibrium position, xf = 0, yf = 0.120 m. using the same initial position as in the first part of the problem,

Ei = ½ kxi2 = Ef = mgyf + ½ mvf

2,

vf = ( (kxi2/m) - 2gyf )1/2

Using m = 0.035 g, k = 953 N/m, yf = 0.120 m, and xi = 0.120 m,

vf = 19.7 m/s

Nonconservative Forces

A nonconservative force is one that does not obey the two conditions for conservative forces. If we take a look at a book being pushed on the surface of a table:

More work will be done moving along path A than along path B, since the direction of the friction force changes to oppose the instantaneous velocity at any instant. Friction acts along a greater distance along path A and thus does more work than along path B. Thus, condition 1 for conservative forces is violated, leaving us with the conclusion that friction in nonconservative.

Resistive forces (friction, air resistance, fluid viscosity) are all nonconservative in nature and will have the effect of removing energy from a system. As the book is moved along the surface of the table, for example, some of the work is converted into a form other than kinetic energy (sound, heat). Thus, the final energy will be less than the initial energy because of this loss of useful work,

It is possible to have a nonconservative force add energy to a system. An applied external force (a person pushing a car up a hill, for example) will add energy to the system if useful work is done by the external force. To account for these possibilities we need to modify our conservation of energy expression.

Conservation of Energy (General Form)

Ef = Ei + WNC

or

½ mvf2 + mgyf + ½ kxf

2 = ½ mvi2 + mgyi + ½ kxi

2 + WNC

where WNC is the work done by nonconservative forces between the initial and final states.

Keep in mind that WNC can be positive or negative. In the case of resistive forces WNC is negative as these forces have the effect of subtracting from a system’s energy.

A crate of mass 3.00 kg slides from rest down the ramp as shown. The length of the ramp is 1.00 m and is inclined at 300 with respect to the horizontal. A constant friction force of 5.00 N opposes the motion of the crate down the ramp. Find the velocity of the crate as it hits the floor.

First we set the bottom of the ramp as the y = 0 position. This makes the y coordinate at the top of the ramp yi = (1.00 m)( sin 300) = 0.5 m.

The best way to solve this type of problem is to first write out expressions for Ei, Ef, and WNC. Noting that vi = 0 and yf = 0,

Ei = mgyi = (3.00 kg)(9.8 m/s2)(0.500 m) = 14.7 J

Ef = ½ mvf2 = ( ½ )(3.00 kg)vf

2

WNC = -fk d = - (5.00 N)( 1.00 m) = - 5.00 J

Ef = Ei + WNC

(1.5 kg)(vf2) = 14.7 J – 5.00 J

vf = (9.7 J/1.5 kg)1/2 = 2.54 m/s

A skier skis down a frictionless ramp from rest from an initial height of 20.0 m. At the bottom of the ramp she skis on a horizontal surface with a coefficient of friction of 0.210. How far does she travel along the horizontal surface before coming to rest?

We will compare the initial energy at A to the final energy at B. Setting the bottom of the ramp as the y = 0 position, we set yf = 0, vi = 0, and vf = 0.

Ei = mgyi

Ef = 0

WNC = -fkd = -μk nd = - μk (mg) d

Ef = Ei + WNC

0 = mgyi - μk (mg) d

d = (yi/μk) = (20 m/0.210) = 95.2 m

A block of mass 0.80 kg is has a velocity of 1.2 m/s on a horizontal frictionless surface. It collides with a spring of negligible mass and compresses an unknown distance before coming to rest and rebounding. If the spring has a force constant of k = 50 N/m find the maximum compression of the spring.

Since the surface is frictionless, WNC = 0. We can set the surface as y = 0, in which case yi = yf = 0. When the spring is fully compressed, the mass comes to rest momentarily, and thus at this point vf = 0. before the mass compresses the spring the spring is fully relaxed ( xi = 0 ). Thus, we can express the individual energy terms as

Ei = ½ mvi2 = ½ (0.8 kg)(1.2 m/s)2 = 0.58 J

Ef = ½ kxf2 = ½ (50 N/m)( xf

2 )

WNC = 0

Ei = 0.58 J = Ef = (25 N/m) xf2

xf = (0.58 J / 25 N/m)1/2 = 0.15 m

Repeat the calculation if a coefficient of friction of μk = 0.50 acts between the surface and the block. Assume that the block has a velocity of 1.2 m/s as it hits the spring.

We have essentially the same problem, only this time we have a nonzero value for WNC. We note that the friction force will act over a distance of xf.

Ei = ½ mvi2 = ½ (0.8 kg)(1.2 m/s)2 = 0.58 J

Ef = ½ kxf2 = ½ (50 N/m)( xf

2 )

WNC = - fk d = -μk n xf = - μk (mg) xf

= - (0.50)(0.8 kg)(9.8 m/s2) xf = (3.92 N) xf

Ef = (25 N/m)(xf2) = Ei + WNC = 0.58 J – (3.92 N) xf

0 = (25 N/m)(xf2) + (3.92 N) xf – 0.58 JThis is a quadratic equation which gives a positive root

solution of

xf = 0.092 m

The negative root can be discarded as this would yield a positive work due to friction.

Homework: # 10, 24, 26, 30, 31, 33, 34, 35, 37.