Chapter

Work and Energy

Blinn College - Physics 2325 - Terry Honan

. - Introduction to Work

Mechanical Advantage

In Chapter D we considered the example of a pulley system lifting a weight.  We saw that using multiple pulleys one can lift a heavy object with a smaller force.  In the example, a weight W could be lifted by a tension T=W/5, but the smaller force must act over a larger distance.  To lift the weight by Δ y one must pull on the rope by Δ x=5 Δ y.  This suggests that force times distance is an important quantity; we will define it as the work.

One Dimensional Work by a Constant Force

If in one dimension we move something by a displacement Δ x with a constant force F.  We will define the work done in this case by

W=F  Δ x.

Note that even in one dimension force and displacement are vector quantities; a one dimensional vector is a real number and the sign gives its direction.  It follows that if the force and displacement are in the same direction then the work is positive and if opposite it is negative.  The work is a scalar quantity.

Units:  The SI unit for work and energy is:  (QuantityUnit)=joule=(QuantityUnit)(QuantityUnit)

Two and Three Dimensional Work by a Constant Force

We now need to generalize this to the case of a constant force (OverVector[F]) and a straight-line path in two or three dimensions.  Here we have a displacement Δ (OverVector[r]) and we will define θ to be the angle between the two vectors  (OverVector[F]) and Δ (OverVector[r]).  What is important is the component of the force along the direction of the displacement, which will be written (Subscript[F, ∥]) and is given by (Subscript[F, ∥])=F cos θ.

The definition of work becomes

W=(Subscript[F, ∥])  Δ r=F Δ r cos θ=F  Δ (Subscript[r, ∥])

where Δ (Subscript[r, ∥]) is the component of Δ (OverVector[r]) parallel to the force (OverVector[F]).

Since (OverVector[F]) and Δ (OverVector[r]) are both vectors and work W is a scalar, the above expressions may be viewed as a product of two vectors giving a scalar.  We will next define the dot or scalar product of two vectors so that the work is written as

W=(OverVector[F])·Δ (OverVector[r]).

. - Vector Algebra II - The Dot Product

The Definition in Terms of Magnitudes and Directions

Given two vectors (OverVector[A]) and (OverVector[B]) we define their Dot or Scalar product to be the scalar quantity given by

(OverVector[A])·(OverVector[B])=A B cos θ,

where θ is the angle between the two vectors.

Properties of the Dot Product

Symmetry

Changing the order of the two vectors doesn't affect the angle between them.  It is clear then that we have the symmetry property :

(OverVector[A])·(OverVector[B])=(OverVector[B])·(OverVector[A]).

Associative Property of Scalar Product with Scalar Multiplication

It doesn't make sense to consider the dot product of more than two vectors, since the dot product of two is no longer a vector.  Associativity is a meaningful thing when combining scalar multiplication with the dot product.  It takes the form:

(c (OverVector[A]))·(OverVector[B])=c ((OverVector[A])·(OverVector[B]))=(OverVector[A])·(c (OverVector[B])).

Distributive Property

When we have both an addition and multiplication operation defined we can ask if the operations are distributive.  Here, there are two ways that the dot product could be distributive and it is distributive both ways

((OverVector[A])+(OverVector[B]))·(OverVector[C])=(OverVector[A])·(OverVector[C])+(OverVector[B])·(OverVector[C])  and  (OverVector[A])·((OverVector[B])+(OverVector[C]))=(OverVector[A])·(OverVector[B])+(OverVector[A])·(OverVector[C]).

Note that mathematically the second expression follows from the first and from the symmetry property.  Combining the distributive and associative properties mentioned shows that the dot product is a linear operation.

The Square of a Vector

Since the angle between a vector and itself is zero and cos 0=1 it follows that for any  (OverVector[A])

(OverVector[A])·(OverVector[A])=(A^2)=((‖ (OverVector[A]) ‖)^2).

Perpendicular Vectors

Since  cos 90°=0 it follows that

(OverVector[A]) ⊥ (OverVector[B])   ⟺   (OverVector[A])·(OverVector[B])=0.

Note that mathematically  "⟹" means implies and "⟺" means that either side implies the other side, or that the two sides are mathematically equivalent.  Here we are defining the zero vector to be perpendicular to everything.

The Dot Product and Components

Using the two results above we can evaluate the dot products of the basis unit vectors   (OverHat[x]),   (OverHat[y]) and  (OverHat[z]).

1=(OverHat[x])·(OverHat[x])=(OverHat[y])·(OverHat[y])= (OverHat[z])· (OverHat[z])  and  0=(OverHat[x])·(OverHat[y])=(OverHat[x])· (OverHat[z])=(OverHat[y])· (OverHat[z])

Using this and the linearity of the dot product we can derive an expression for the dot product in terms of components.

({{(OverVector[A])·(OverVector[B])=((Subscript[A, x]) (OverHat[x])+(Subscript[A, y]) (OverHat[y])+(Subscript[A, z]) (OverHat[z]))·((Subscript[B, x]) (OverHat[x])+(Subscript[B, y]) (OverHat[y])+(Subscript[B, z]) (OverHat[z]))
= (Subscript[A, x]) (Subscript[B, x]) (OverHat[x])·(OverHat[x])+(Subscript[A, x]) (Subscript[B, y]) (OverHat[x])·(OverHat[y])+(Subscript[A, x]) (Subscript[B, z]) (OverHat[x])· (OverHat[z]) 
+(Subscript[A, y]) (Subscript[B, x]) (OverHat[y])·(OverHat[x])+(Subscript[A, y]) (Subscript[B, y])  (OverHat[y])·(OverHat[y])+(Subscript[A, y]) (Subscript[B, z]) (OverHat[y])· (OverHat[z])
+(Subscript[A, z]) (Subscript[B, x])  (OverHat[z])·(OverHat[x])+(Subscript[A, z]) (Subscript[B, y])  (OverHat[z])·(OverHat[y])+(Subscript[A, z]) (Subscript[B, z])  (OverHat[z])· (OverHat[z])}})

Using the expressions above for the dot products of unit vectors we get all the off-diagonal terms going to zero and the dot products in the diagonal terms giving one.  We end up with the result

(OverVector[A])·(OverVector[B])=(Subscript[A, x]) (Subscript[B, x])+(Subscript[A, y]) (Subscript[B, y])+(Subscript[A, z]) (Subscript[B, z]).

This will be referred to as the component form of the dot product and should be viewed as an alternative definition.

Example . - Angle Between Two Vectors

(OverVector[A])=⟨3,5,-4⟩   and   (OverVector[B])=⟨-3,2,1⟩

Solution

(OverVector[A])·(OverVector[B])=(Subscript[A, x]) (Subscript[B, x])+(Subscript[A, y]) (Subscript[B, y])+(Subscript[A, z]) (Subscript[B, z])=3 (-3)+5·2+(-4) 1=-3

A=‖(OverVector[A])‖=(Sqrt[(Subscript[A, x]^2)+(Subscript[A, y]^2)+(Subscript[A, z]^2)])=(Sqrt[(3^2)+(5^2)+(((-4))^2)])=(Sqrt[50])   and

B=‖(OverVector[B])‖=(Sqrt[(Subscript[B, x]^2)+(Subscript[B, y]^2)+(Subscript[B, z]^2)])=(Sqrt[(((-3))^2)+(2^2)+(1^2)])=(Sqrt[14])

(OverVector[A])·(OverVector[B])=A B cos θ   ⟹   θ=(cos^(-1))((((OverVector[A])·(OverVector[B]))/(A B)))=96.5°

Example . - Dragging a Crate

Solution

T=115 (QuantityUnit), f=80 (QuantityUnit), m=38 (QuantityUnit), θ=35°  and  Δ x=4 (QuantityUnit)

W=(OverVector[F])·Δ (OverVector[r])=F Δ x cos θ.

(Subscript[W, T])=T Δ x cos θ=377.(QuantityUnit)

(Subscript[W, f])=f Δ x cos 180°=-f Δ x=-320(QuantityUnit)

Solution

(Subscript[F, net,hor])=T cos θ-f=m a  ⟹  a=(1/m)(T cos θ-f)=0.374(((QuantityUnit))/(((QuantityUnit))^2))

(v^2)=(Subscript[v, 0]^2)+2 a Δ x   ⟹   v=(Sqrt[2 a Δ x ])=1.73(QuantityUnit)/(QuantityUnit)

. - Work in General

So far we have the definition of work for a constant force and a straight line path to be W=(OverVector[F])·Δ (OverVector[r]).  We need to generalize this to a varying force acting on a general path that is allowed to curve. First let us consider an infinitesimal segment of the path  d(OverVector[r]). Over a sufficiently small segment the path is straight and the force varies negligibly. It follows that the work over the infinitesimal segment is  (OverVector[F])·d(OverVector[r]).  A definite integral is an infinite sum over infinitesimal pieces.  We denote this infinite sum with the integral symbol "∫" and write the work as

W=∫(OverVector[F])·d(OverVector[r]).

The integral in this definition of work is of a vector field integrated along a general path.  This is a mathematical object that a student will learn to evaluate in the third semester of the calculus sequence.  We will not evaluate these general integrals this semester.  We will discuss the integrals in general and only evaluate special cases.  Some of these special cases will be discussed below.

Constant Force

If a force is constant then we can take it out of the above general integral.  This leaves the integral ∫d(OverVector[r]) which is just the sum over all the infinitesimal pieces  d(OverVector[r]); this becomes the vector Δ (OverVector[r]) which is the vector from the starting position to the end of the path.

W=∫(OverVector[F])·d(OverVector[r])=(OverVector[F])·∫d(OverVector[r])=(OverVector[F])·Δ (OverVector[r]).

Note that this is the same result we had for a constant force over a straight-line path

Work Done by Gravity

An important special case of the previous result is the work done by gravity.  The force is a uniform  (OverVector[F])=-m g  (OverHat[y]) and the work becomes

(Subscript[W, grav])=(OverVector[F])·Δ (OverVector[r])=-m g  (OverHat[y])·Δ (OverVector[r])=-m g Δ y.

To derive this we used that  (OverHat[y])·Δ (OverVector[r]) is the y component of Δ (OverVector[r]), which is Δ y.

Example . - A Box on a Table

Solution

(Subscript[W, grav])=-m g Δ y=m g h.

Solution

(Subscript[W, grav])=-m g Δ y=-m g h.

Solution

(Subscript[W, grav])=-m g Δ y=0.

One Dimensional Work by a Varying Force

If we have F(x) as a one dimension force as a function of position.  The work when moving from position (Subscript[x, i]) to (Subscript[x, f]) the work is the one dimensional definite integral

W=(∫_(Subscript[x, i])^(Subscript[x, f])) F(x) dx.

This is the sort of one variable integral discussed in the first semester of calculus and the student will be responsible for such integrals.  To evaluate a definite integral one uses the fundamental theorem of calculus.

The Fundamental Theorem of Calculus

The fundamental theorem of calculus relates integral and differential calculus.  It is the rule one uses to evaluate a definite integral.

(∫_a^b) g(x) dx=G(x)(Subscript[|, a]^b)=G(b)-G(a)  where  G'(x)=g(x)

The definite integral of a function is the difference of some antiderivative of the function evaluated at the endpoints of the interval.

Example . - Work Done by a Varying Force

F(x)=18x-6(x^2)+4   (in SI units).

Solution

({{W=(∫_(Subscript[x, i])^(Subscript[x, f])) F(x) dx=(∫_(1 (QuantityUnit))^( 4 (QuantityUnit))) (18x-6(x^2) +4) dx}, {(Subscript[=(9(x^2)-2(x^3)+4x)|, 1 (QuantityUnit)]^(4 (QuantityUnit)))=(144-128+16)(QuantityUnit)-(9-2+4)(QuantityUnit)}, {=21(QuantityUnit)}})

Hooke's Law and Elastic Forces

Hooke's law is the force law for a spring.  Define x=0 to be the relaxed position of a spring, the equilibrium position, and let x be the amount the spring is stretched from equilibrium.  Hooke's law states that there is a proportionality between F and x, F∝x.  We can introduce a constant of proportionality k called the spring constant or force constant.

F=k x  (Force on spring)

The spring constant is a property of a particular spring; a stiff spring has a large k and a loose spring has a small k.  In this expression F is the force stretching the spring.  Usually we consider the force of the spring itself when considering Hooke's law.  By Newton's third law this is the negative of the force on the spring and we get

F=-k x  (Force of spring).

Note that when the spring is stretched, x is positive and the force is negative.  When the spring is compressed the force is positive and x is negative.

A force is said to be elastic when it satisfies Hooke's law.  We will see that elasticity is quite common for sufficiently small deformations of almost anything.  What is special about springs is they maintain their elasticity over large deformations.  It should be clear that even a spring will violate Hooke's law when stretched a very large distance.

Work Done by a Spring

We may now derive an expression for the work done by a spring.  Using the formula for the work done in one dimension by a varying force  F(x) and Hooke's law  F(x)=-k x we get a simple integral.

(Subscript[W, spring])=(∫_(Subscript[x, i])^(Subscript[x, f])) F(x) dx=-k (∫_(Subscript[x, i])^(Subscript[x, f])) x dx.

Using the fact that (1/2)(x^2) is the antiderivative of x we get

(Subscript[W, spring])=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2)).

Example . - Hooke’s Law

Solution

F=60(QuantityUnit)   and   x=0.04(QuantityUnit)

F=k x   ⟹   k=F/x=1500(QuantityUnit)/(QuantityUnit).

Solution

(Subscript[W, spring])=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2))=-(1/2)k ((x^2)-(0^2))=-(1/2)k (x^2)=-1.2(QuantityUnit) .

. - Power

Power, in the most general sense, is the rate that something uses or provides energy.

𝒫=((dEnergy)/(dt)).

The power delivered by a motor or engine is the rate that it can do work

𝒫=((dW)/(dt)).

If we write the work in terms of the force we get  dW=(OverVector[F])·d(OverVector[r]).  Writing the infinitesimal displacement as  d(OverVector[r])=(OverVector[v]) dt we get

𝒫=(OverVector[F])·(OverVector[v]).

Units:  The SI unit for power is:  (QuantityUnit)=watt=(QuantityUnit)/(QuantityUnit)

. - Kinetic Energy and the Work-Energy Theorem

The Net Work

Newton's second law isn't a general statement about forces but is about the net force acting on a body,  (Subscript[(OverVector[F]), net])=m (OverVector[a]).  (Subscript[(OverVector[F]), net]) is the vector sum of all forces acting on a body; we use a free-body diagram to help us sum these forces.  Let us symbolically write the net force in terms of all the forces acting on a body (all the forces in the free-body diagram) as

(Subscript[(OverVector[F]), net])=(Subscript[(OverVector[F]), 1])+(Subscript[(OverVector[F]), 2])+....

Each force acting on a body does work on the body.  (Some of these works may be zero, however.)  If the work done by (Subscript[(OverVector[F]), i]) is labeled (Subscript[W, i]) then we can define the net work as the sum of all these works.

(Subscript[W, net])=(Subscript[W, 1])+(Subscript[W, 2])+....

The Work-Energy Theorem

The work-energy theorem is a very important result.  It is where the idea of energy comes into physics and it explains why work is a useful notion.  We define the kinetic energy by

K=(1/2)m (v^2).

The theorem is quite simple to state; it equates the net work to the change in the kinetic energy.

(Subscript[W, net])=Δ K

where the change is kinetic energy is  Δ K=(Subscript[K, f])-(Subscript[K, i])=(1/2)m  ((Subscript[v, f]^2) - (Subscript[v, i]^2)).

Example . - Dragging a Crate (continued)

(Subscript[W, T])=376.8(QuantityUnit) ,  (Subscript[W, f])=-320.0(QuantityUnit)  and (Subscript[W, N])=0=(Subscript[W, grav])

Solution

(Subscript[W, net])=(Subscript[W, T])+(Subscript[W, f])+(Subscript[W, N])+(Subscript[W, grav])=376.8(QuantityUnit)-320.0(QuantityUnit)+0+0=56.8(QuantityUnit)

Solution

(Subscript[W, net])=Δ K=(1/2)m ((Subscript[v, f]^2)-(Subscript[v, i]^2))

(Subscript[v, f])=(Sqrt[(2/m)(Subscript[W, net])])=1.73(QuantityUnit)/(QuantityUnit)

Proof of the Theorem

Consider a single mass moving along a general path.  The net work is the sum of the works done by each force acting on a body.  Since the integral of the sum is the sum of the integrals, the net work can be written as the work done by the net force.

(Subscript[W, net])=∫(Subscript[(OverVector[F]), net])·d(OverVector[r]).

We then apply the second law (Subscript[(OverVector[F]), net])=m (OverVector[a]) and use the definition of velocity to write d(OverVector[r])=(OverVector[v]) dt.  We then get

(Subscript[W, net])=∫m (OverVector[a])·(OverVector[v]) dt.

We have now turned the integral over a contour into a simple integral over a single variable.  The limits of this integration are the initial and final times.

(Subscript[W, net])=(∫_( (Subscript[t, i]))^(Subscript[t, f])) (m (OverVector[a])·(OverVector[v])) dt.

To evaluate the above integral we need to, using the fundamental theorem of calculus, find the antiderivative of the integrand  m (OverVector[a])·(OverVector[v]).  This antiderivative is just the kinetic energy:

(d/(dt))((1/2)m (v^2))=m (OverVector[a])·(OverVector[v]).

To verify this, note that dot products satisfy the usual product rule:

(OverVector[A])·(OverVector[B])=(Subscript[A, x]) (Subscript[B, x])+(Subscript[A, y]) (Subscript[B, y])+(Subscript[A, z]) (Subscript[B, z])  ⟹  (d/(dt))((OverVector[A])·(OverVector[B]))=((d/(dt))(OverVector[A]))·(OverVector[B])+(OverVector[A])·((d/(dt))(OverVector[B]))

and thus

(d/(dt)) (v^2)=(d/(dt))(OverVector[v])·(OverVector[v])=2 ((d/(dt))(OverVector[v]))·(OverVector[v])=2 (OverVector[a])·(OverVector[v]).

It is now clear that the kinetic energy is the antiderivative of  m (OverVector[a])·(OverVector[v]).

Using the fundamental theorem of calculus we get the work energy theorem

(Subscript[W, net])=(∫_( (Subscript[t, i]))^(Subscript[t, f])) (m (OverVector[a])·(OverVector[v])) dt=(1/2)m ((v((Subscript[t, f])))^2)-(1/2)m ((v((Subscript[t, i])))^2)=Δ K.

. - Conservative Forces and Potential Energy

When a body is moved in a uniform gravitational field (OverVector[g]) the work done by gravity is given by: W=- m g  Δ y Note that if the body is moved along different paths with the same endpoints (starting and stopping points) the Δ y is the same and thus the work done by gravity is the same.  In stretching a spring the work of the spring W=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2))  depends only on the endpoints (Subscript[x, i]) and (Subscript[x, f]) and not on the details of the path.

We will define such forces (where the work is independent of path) as conservative and we will then be able to define a new type of energy, called potential energy or U, for these forces.  The entire effect of the work of these forces will be incorporated into considering the potential energy functions at the endpoints of the paths.  Potential energy will be an easy and useful bookkeeping tool for keeping track of the work contributions for conservative forces in the work-energy theorem.

Conservative and Nonconservative Forces

A force is defined to be a conservative force when its work is independent of the path taken.  Equivalently, we can say that for a conservative force the work around any closed path (a path that ends where it begins) is zero.

Putting a circle in the integral sign implies that the path is closed and the definition of conservative is written:

∮ (OverVector[F])·d(OverVector[r])=0  ⟺   (OverVector[F]) is conservative.

Conservative Examples

Examples of conservative forces have already been mentioned.  These are uniform gravitational forces and the elastic force of a spring.  Other examples of conservative forces that will be encountered are nonuniform gravitational forces, which will be discussed later this semester, and the electrostatic force, which will be considered in the second semester course.

Dissipative friction is nonconservative.

Consider an object being dragged along a horizontal floor by a horizontal rope.  If the dragging force is horizontal, the normal force is the weight and the magnitude of the force of kinetic friction is (Subscript[f, k])=(Subscript[μ, k]) m g = constant.  The direction of the force of friction opposes the direction of motion so the work done by friction becomes

(Subscript[W, f])=∫ (Subscript[(OverVector[f]), k])·d(OverVector[r])=-(Subscript[f, k]) ∫ds.

Note that ds is the magnitude of the infinitesimal displacement d(OverVector[r]),  ds=‖d(OverVector[r])‖, and is then the infinitesimal arc length.  The integral ∫ds is then the total arc length.  It is clear that (Subscript[W, f]) depends on the path length and that kinetic friction isn't conservative.

Driving forces are nonconservative.

By driving forces we mean the force of a car's engine propelling a car or the force of a cyclist propelling a cycle.  Consider a car that begins at some location, drives on some path and then returns to the same point.  The car's engine does work in this process and thus cannot be conservative.  A conservative force must give zero work for a closed path.

Other Nonconservative Forces

Any force that is not conservative is nonconservative, where our two conservative examples are gravity or the elastic force of a spring. If a tension force or normal force acts on a body then those also represent nonconservative forces; in some examples the tension and normal force will be present but do no work. If there is some external applied force acting on a body then that is nonconservative.

Potential Energy

For any conservative force we can define a potential energy, U.  This idea is this: since the work depends only on the endpoints of a path and not the details of the path then we can write the work as the difference of some function evaluated only at the endpoints.  We will define this function as the negative of the potential energy function.  The reason for the sign will become clear later.

The definition of potential energy is

Δ U=-W=-∫(OverVector[F])·d(OverVector[r]).

The zero of potential energy is arbitrary.  In some cases there will be standard choices of the zero position.

Potential Energy for Uniform Gravity

Since for gravity we have W=-m g Δ y, we define gravitational potential energy by Δ U=m g Δ y.  We can choose the zero of potential energy to be where y=0 and then define the potential energy function as

U=m g y.

The zero of y is still arbitrary.

Elastic Potential Energy

The work done by a spring is given by W=-(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2)).  When we take  Δ U=-W we get

Δ U=-W=(1/2)k ((Subscript[x, f]^2)-(Subscript[x, i]^2)).

We want to find a function U(x) that satisfies Δ U=U((Subscript[x, f]))-U((Subscript[x, i])).  The easiest choice is

U=(1/2)k (x^2).

In making this choice we take the zero position of potential energy to be the equilibrium position x=0.

Work and Mechanical Energy

Let us now apply these ideas to the work-energy theorem.  We begin by writing all forces acting on a body as

(Subscript[(OverVector[F]), net])=(Underscript[(Underscript[   (Subscript[(OverVector[F]), nc]) , ⏟])   , ({{all nonconservative}, {forces}})])+  (Underscript[(Underscript[(Subscript[(OverVector[F]), 1])+(Subscript[(OverVector[F]), 2])+..., ⏟]), conservative forces]) .

(Subscript[(OverVector[F]), nc]) represents the sum of all nonconservative forces.  The other forces are the conservative forces listed separately.  We now make the same decomposition of the corresponding works.

(Subscript[W, net])=(Subscript[W, nc])+(Subscript[W, 1])+(Subscript[W, 2])+...

Now we make the replacements  (Subscript[W, i])=-Δ (Subscript[U, i]).  Plugging the above expression for (Subscript[W, net]) into the work-energy theorem (Subscript[W, net])=Δ K and moving the Δ (Subscript[U, i]) terms to the right hand side gives:

(Subscript[W, nc])=Δ K+Δ (Subscript[U, 1])+Δ (Subscript[U, 2])+....

This result applies to a single mass.  To apply it to a system containing multiple masses, like for instance Atwood's machine, we can sum this over every mass in the system.  Now take (Subscript[W, nc]) to be the sum of the (Subscript[W, nc]) for all the masses.  Call  (Subscript[K, tot]) the sum of the kinetic energies of all masses and (Subscript[U, tot]) the sum over all the (Subscript[U, i]) for all the masses.  We end up with the result

(Subscript[W, nc])=Δ (Subscript[K, tot])+Δ (Subscript[U, tot])
=Δ (Subscript[E, mech])

where we have defined the total mechanical energy of the system as  (Subscript[E, mech])=(Subscript[K, tot])+(Subscript[U, tot]).  Usually the mechanical energy will just be written as E.

Nonconservative forces will usually consist of friction forces, which remove mechanical energy from a system, and driving forces like the work done by car's engine or a cyclist, which add mechanical energy.

. - Conservation of Energy

There are two notions of conservation of energy we will consider.  One will be of practical importance for problem solving.  The other is a very fundamental notion.

Conservation of Mechanical Energy

The conservation of mechanical energy is a principle that will prove very useful in problem solving.  Begin with the fundamental expression  (Subscript[W, nc])=Δ (Subscript[E, mech]).  If in some problem there are no nonconservative forces (i.e. no dissipative friction or driving force) then we can conclude that Δ (Subscript[E, mech])=0 or that

(Subscript[E, mech,i])=(Subscript[E, mech,f])  or   (Subscript[E, i])=(Subscript[E, f]).

To solve such a problem we need to find an expression for E, the total mechanical energy.  To do this we add a kinetic energy for each mass in the problem, add in gravitational potential energies for each mass and add an elastic potential energy for each spring.

Energy as a Fundamentally Conserved Quantity

Energy is a fundamentally conserved quantity.  This means that it cannot be created or destroyed; we can just convert it from one form to another.

Consider the mechanical energy of a car  (Subscript[W, nc])=Δ (Subscript[E, mech]).  The mechanical energy is not conserved due to the (Subscript[W, nc]) of friction and the engine.  The energy lost to friction goes into heat.  The energy from the engine comes from the energy stored in the chemical bonds of the fuel.

When solving problems that involve nonconservative forces we can rewrite (Subscript[W, nc])=Δ (Subscript[E, mech])=(Subscript[E, f])-(Subscript[E, i]) as

(Subscript[E, i])+(Subscript[W, nc])=(Subscript[E, f])

Written this way, problem solving is more similar for both types of problems, where (Subscript[W, nc]) is zero or not.

Example . - A Rolling Car

Solution

m=1500 (QuantityUnit),  h=12 (QuantityUnit)  and  (Subscript[v, i])=21 (QuantityUnit)/(QuantityUnit).

E=(1/2)m (v^2)+m g y   and   (Subscript[E, i])=(Subscript[E, f])  ⟹  (1/2)m (Subscript[v, i]^2)+0=(1/2)m (Subscript[v, f]^2)+m g h

(Subscript[v, f])=(Sqrt[(Subscript[v, i]^2)-2 g h])=14.3(QuantityUnit)/(QuantityUnit)

Solution

(Subscript[W, nc])=(Subscript[W, friction])+(Subscript[W, engine])=(Subscript[W, friction])+0

E=(1/2)m (v^2)+m g y   and   (Subscript[E, i])+(Subscript[W, nc])=(Subscript[E, f])  ⟹  (1/2)m (Subscript[v, i]^2)+0+(Subscript[W, friction])=(1/2)m (Subscript[v, f]^2)+m g h

(Subscript[W, friction])=(1/2)m (Subscript[v, f]^2)+m g h-(1/2)m (Subscript[v, i]^2)=-86700(QuantityUnit)

Example . - A Spring Gun

Solution

E=(1/2)m (v^2)+m g y+(1/2)k (x^2).

(Subscript[E, i])=(Subscript[E, f])  ⟹  0+m g h+(1/2)k (Subscript[x, 0]^2)=(1/2)m (Subscript[v, f]^2)+0+0

(Subscript[v, f])=(Sqrt[(k/m) (Subscript[x, 0]^2)+2 g h]).

Example . - A Spring and a Block

Solution

m=4 (QuantityUnit),  k=800(QuantityUnit)/(QuantityUnit), (Subscript[x, 0])=0.10(QuantityUnit)  and  d=0.30(QuantityUnit)

E=(1/2)m (v^2)+(1/2)k (x^2)

(Subscript[W, nc])=(Subscript[W, friction])=-(Subscript[f, k])Δ x=-(Subscript[μ, k])N Δ x=-(Subscript[μ, k])m g ((Subscript[x, 0])+d)

(Subscript[E, i])+(Subscript[W, nc])=(Subscript[E, f])  ⟹  0+(1/2)k (Subscript[x, 0]^2)-(Subscript[μ, k])m g ((Subscript[x, 0])+d)=0+0

(Subscript[μ, k])=((k (Subscript[x, 0]^2))/(2 m g ((Subscript[x, 0])+d)))=0.255 .

Example . - Two Connected Masses

Solution

K=(1/2)(Subscript[m, 1])(Subscript[v, 1]^2)+(1/2)(Subscript[m, 2])(Subscript[v, 2]^2)=(1/2)(Subscript[m, 1])(v^2)+(1/2)(Subscript[m, 2])(v^2)=(1/2)((Subscript[m, 1])+(Subscript[m, 2]))(v^2)

E=(1/2)((Subscript[m, 1])+(Subscript[m, 2]))(v^2)+(Subscript[m, 2]) g (Subscript[y, 2])

(Subscript[E, i])=(Subscript[E, f])   ⟹   0+(Subscript[m, 2]) g h=(1/2)((Subscript[m, 1])+(Subscript[m, 2]))(Subscript[v, f]^2)+0

(Subscript[v, f])=(Sqrt[2 ((((Subscript[m, 2])g)/((Subscript[m, 1])+(Subscript[m, 2])))) h])

Solution

(Subscript[W, nc])=(Subscript[W, friction])=-(Subscript[f, k])Δ x=-((Subscript[μ, k])N) Δ x=-((Subscript[μ, k])(Subscript[m, 1])g) h

(Subscript[E, i])+(Subscript[W, nc])=(Subscript[E, f])  ⟹  (0+(Subscript[m, 2]) g h)-(Subscript[μ, k])(Subscript[m, 1])g h=(1/2)((Subscript[m, 1])+(Subscript[m, 2]))(Subscript[v, f]^2)+0

(Subscript[v, f])=(Sqrt[2 ((((Subscript[m, 2])g-(Subscript[μ, k])(Subscript[m, 1])g)/((Subscript[m, 1])+(Subscript[m, 2])))) h])

. - Force, Potential Energy and Stability

The basic relation between potential energy and force is

Δ U=-∫(OverVector[F])·d(OverVector[r]).

This is the rule for going from the force as a function of position to the potential energy.  If we move along some path, the path of integration, then we get Δ U for the change in potential energy along that path.  The infinitesimal form of this is

dU=-(OverVector[F])·d(OverVector[r]).

This is the infinitesimal change in the potential energy when one moves along the infinitesimal displacement d(OverVector[r]).

Force from Potential Energy

The rule for going from force to potential energy leads to the question: can we do this in reverse and go from the potential energy to the force?  The answer is yes.  To get the x component of the force consider an infinitesimal displacement in the x direction  d(OverVector[r])=(OverHat[x]) dx.  Using the fact that (OverVector[F])·(OverHat[x])=(Subscript[F, x]) we get

dU=-(OverVector[F])·d(OverVector[r])=-(OverVector[F])·(OverHat[x]) dx=-(Subscript[F, x]) dx.

It is then a simple matter to solve for the x component of the force

(Subscript[F, x]) =-((dU)/(dx)).

This is simple but it is too simple, to the point of being incorrect when the potential energy is a function of more than one variable.  For functions of more than one variable we take partial derivatives and not ordinary derivatives.

(Subscript[F, x]) =-((∂U)/(∂x)),  (Subscript[F, y]) =-((∂U)/(∂y))  and  (Subscript[F, z]) =-((∂U)/(∂z)).

To evaluate a partial derivative take the derivative with respect to the variable treating the other independent variables as constants.  In a case when the potential energy is only a function of one variable the original expression with the ordinary derivative is correct, but we do not need the component subscript

F =-((dU)/(dx)).

As an example consider gravity.  Take the potential to be a function of a horizontal and a vertical variable, x and y.

U(x,y)=m g y      ⟹      
(Subscript[F, x]) =-((∂U)/(∂x))=0   and   (Subscript[F, y]) =-((∂U)/(∂y))=-m g

This is the proper expression for the force.  For another example consider the one dimensional case of the elastic potential energy of a spring.

U(x)=(1/2)k (x^2)  ⟹  F =-((dU)/(dx))=-k x

This is our familiar Hooke's law expression.

Example . - Partial Derivatives and Potential Energy

f(x,y)=a (x^3)y-b (y^2)

Solution

((∂f)/(∂x))=3a (x^2)y   and   ((∂f)/(∂y))=a (x^3)-2b y

U(x,y)=a (x^3)y-b (y^2)

Solution

(Subscript[F, x])=-((∂U)/(∂x))=-3a (x^2)y   and   (Subscript[F, y])=-((∂U)/(∂y))=-a (x^3)+2b y

(OverVector[F])=⟨(Subscript[F, x]),(Subscript[F, y])⟩=⟨-3a (x^2)y ,-a (x^3)+2b y⟩

Stability and Potential Energy Diagrams

The expression  F=-dU/dx implies that the direction of the force is always toward lower potential energy.  If we consider a graph of potential energy as a function of position then the force is toward lower U.  For instance is the slope of U vs. x is positive then the force is negative and if the slope is negative the force is positive.

An equilibrium point is a point of zero force and thus zero slope.  These are the local minima and maxima.  If x is a small distance away from a local minima then the force is toward the minimum; we call this a stable equilibrium.  Around a local maximum the opposite happens.  When a small distance away from equilibrium the force is away; this is called an unstable equilibrium.