Chapter

Potential Energy and Energy Conservation

Blinn College - Physics 1401 - Terry Honan

. - 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; for any two paths with the same endpoints (Subscript[W, 1])=(Subscript[W, 2]). Equivalently, we can say that for a conservative force the work around any closed path (a path that ends where it begins) is zero.

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 for a small displacement d is -(Subscript[f, k])d, since cos 180°=-1. For a long path, summing over all then small displacements gives

(Subscript[W, f])=-(Subscript[f, k]) (path length)

Since it depends on the path length, it is clear that kinetic friction is not 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

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])