Chapter 7 Potential Energy and Energy Conservation 7.1

Gravitational Potential Energy

So far the only form of energy that we defined was kinetic energy which is a function of velocity squared (or speed squared) 1 K = mv 2 . 2

(7.1)

It turns out that it is also useful to define another form of energy which is a function of position. This new physical quantity we will call potential energy. The idea behind name is that it might be possible to store energy in the form of potential energy by placing an object at certain position. The process of storing potential energy would most certainly require work to be done on the object, but later on the potential energy can be released. The most familiar example of potential energy is the gravitational potential energy, when the work must be done on an object to lift it up. For example, if on object has mass m and is lifted from height x1 to height x2 , then the work done by gravitational force is ! "! " Wgrav = −mg ˆi (x2 − x1 ) ˆi = mgx1 − mgx2 (7.2) Then it is convenient to define a gravitational potential energy Ugrav ≡ mgx

(7.3)

and then the work done by gravitational force can be written as Wgrav = −∆Ugrav = Ugrav,1 − Ugrav,2 . 78

(7.4)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 79 Now if we assume that there are no other forces acting on an object, then according to work-energy theorem Wgrav = K2 − K1

(7.5)

and by combining (7.4) and (7.5) we get Ugrav,1 − Ugrav,2 = K2 − K1

(7.6)

Ugrav,1 + K1 = Ugrav,2 + K2 .

(7.7)

or The latter equation suggest that the total (mechanical) energy is unchanged E ≡ K + Ugrav = constant.

(7.8)

More generally there might be other forces acting on a given system which might do work to it and then the work-energy theorem would imply Wother = Ef − Ei

(7.9)

Note that although the conservation of energy suggests that the total energy of all systems should still be conserved, the mechanical energy of any one system changes according to (7.9). Conservation of energy is the first of many conservation laws. As well as any other conservation law, the conservation of energy is due to certain symmetries (in this case time shift symmetry) that play a central role in modern physics. Example 7.1 . You throw a 0.145 − kg baseball straight up, giving it an initial velocity of magnitude 20.0 m/s. How high it goes?

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 80 Step 1: Coordinate system. It is sufficient to choose a 1D coordinate system with x-axis pointing upwards and origin at the initial position of the ball. Step 2: Initial conditions. The initial velocity is ⃗vi = 20.0 m/s ˆi,

(7.10)

1 Ki = mvi2 2

(7.11)

Ui = mgxi = 0 J.

(7.12)

initial kinetic energy is

and initial potential energy is

Step 3: Final conditions. Then from conservation of energy Uf + Kf = Ui + Ki 1 mgxf + 0 = 0 + mvi2 2 and thus xf =

(20.0 m/s)2 vi2 = # $ = 20.4 m. 2g 2 9.8 m/s2

(7.13)

(7.14)

Note that we did not have to know the mass of the baseball. Example 7.2. In Example 7.1 suppose your hand moves upward by 0.50 m while you are throwing the ball. The ball leaves your hand with an upward velocity of 20 m/s. (a) Find the magnitude of the force (assumed constant) that your hand exerts on the ball. (b) Find the speed of the ball at a point 15.0 m above the point where it leaves your hand. Ignore air resistance. Step 1: Coordinate system. We can choose the same coordinate system as in Example 7.1. Step 2: Free body diagram. There are two phases of motion, but we only need the free-body diagram for the first phase with origin at the initial position of the ball.

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 81

Initial condition for second phase / final condition for first phase: Velocity is

⃗v2 = 20.0 m/s ˆi,

(7.15)

1 1 K2 = mv22 = (0.145 kg) (20.0 m/s)2 = 29.0 J 2 2

(7.16)

kinetic energy is

potential energy is # $ U2 = mgx2 = (0.145 kg) 9.8 m/s2 (0.50 m) = 0.71 J.

(7.17)

total mechanical energy is

E2 = K2 + U2 = 29.7 J

(7.18)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 82 Step 3: Apply Newton’s Laws / Work -energy theorem. (a) For the first phase the (generalized) work-energy theorem implies Wother = E2 − E1 F x2 = E2 E2 29.7 J F = = = 59 N. x2 0.50 m

(7.19)

(b) For the second phase the work-energy theorem implies E3 K3 + U3

= =

E2 E2

1 2 mv + mgx3 2 3

=

E2 %

v3 = ± or

2 (E2 − mgx3 ) m

(7.20)

& # # $ $ 2 59 N − (0.145 kg) 9.8 m/s2 (0.50 m + 15.0 m) v3 = = 10 m/s. (7.21) (0.145 kg) Note that speed must be a positive number and so we only picked a positive solution from (7.20). Example 7.3. A batter hits two identical baseballs with the same initial speed and from the same initial height but at different initial angles. Prove that both balls have the same speed at any height h is air resistance can be neglected.

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 83 The total mechanical energy at the initial conditions is the same in both cases, and from conservation of energy the mechanical energy at any moment of time is also the same in both cases. Moreover when balls are at the same hight the potential energies are the same and therefore the kinetic energies and speeds are the same. U1i + K1i E1i E1f U1f + K1f K1f v1f

= = = = = =

U2i + K2i from the same initial speed and height E2i from definition of total energy E2f from conservation of energy for each trajectory U2f + K2f from definition of total energy K2f from definition of potential energy v2f from definition of kinetic energy (7.22)

Example 7.6. We want to slide a 12 − kg crate up a 2.5 − m-long ramp inclined at 30◦ angle. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom of and letting it go. But, friction is not negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. (a) Find the magnitude of friction acting on crate, assuming that it is constant. (b) How fast is the crate moving when it reaches the bottom of the ramp?

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 84

Step 1: Coordinate system. Let’s choose x-axis to point horizontally and y-axis to point vertically with origin at the initial state. Step 2: Initial conditions. At the initial conditions velocity is ⃗v1 = 5.0 m/s ˆi

(7.23)

1 1 K1 = mv12 = (12 kg) (5.0 m/s)2 = 150 J 2 2

(7.24)

U1 = mgy1 = 0

(7.25)

E1 = U1 + K1 = 150 J.

(7.26)

kinetic energy

potential energy and total energy is Step 3: Apply the work-energy theorem. There are two stages of motion. In

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 85 the first stage the work energy theorem implies W1−2 = E2 − E1 −F s = U2 + K2 − E1 # $ −F (1.6 m) = (12 kg) 9.8 m/s2 (1.6 m) sin (30◦ ) + 0 − 150 J 56 J F = = 35 N. (7.27) 1.6 m where F is the magnitude of the force of friction and −F s is the work done by friction which is always a negative number. In the second stage W2−3 = E3 − E2 −F s = U3 + K3 − U2 − K2 # $ 1 − (35 N) (1.6 m) = 0 + (12 kg) v32 − (12 kg) 9.8 m/s2 (1.6 m) sin (30◦ ) − 0 & 2 v3 =

2 (94 J − 56 J) = 2.5 m/s. 12 kg

(7.28)

where v3 is velocity of the crete when it reached the bottom of crate.

7.2

Elastic Potential Energy

Remember that the work done on a spring is given by 1 1 W = kx2f − kx2i 2 2

(7.29)

but so the work done by a spring is 1 1 Wel = kx2i − kx2f 2 2

(7.30)

Similarly to work done by gravity, Wgrav = −∆Ugrav = mgyi − mgyf

(7.31)

the work done by a spring can be thought as a difference of the potential energies 1 1 (7.32) Wel = −∆Uel = kx2i − kx2f 2 2 with the difference that Ugrav = mgy (7.33)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 86 but

1 (7.34) Uel = kx2 2 where it is assumed that the origin is chosen at a position corresponding to unstretched spring. It is sometime useful to combine both potential energies together, to define a total potential energy 1 U = mgy + kx2 2

(7.35)

(where depending on the setup x and y may or may not be the same) and also total mechanical energy as E =U +K

(7.36)

We can now modify the work-energy theorem further by stating that if there are other forces (not gravitational, nor elastic) doing work on the system then the total mechanical energy of the system changes according to Wother = Ef − Ei .

(7.37)

Example 7.7. A glider with mass m = 0.200 kg sits on a frictionless horizontal air track, connected to a spring with constant k = 5.00 N/m. You pull on the glider, stretching the spring 0.100 m, and releasing it from rest. The glider moves back towards its equilibrium position (x = 0). What is its x-velocity when x = 0.080 m?

Step 1: Coordinate system. Let’s choose x-axis to point to the right with origin corresponding to equilibrium of spring. Step 2: Initial conditions. At the initial conditions the velocity ⃗vi = 0

(7.38)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 87 kinetic energy

(elastic) potential energy

1 Ki = mvi2 = 0 2

(7.39)

1 Ui = kx2i 2

(7.40)

Ei = Ui + Ki = Ui

(7.41)

and total energy Step 3: Apply work-energy theorem. From the conservation of energy Ef = Ei Uf + Kf = Ui 1 2 1 2 1 2 kxf + mvf = kx 2 2 2 i

(7.42)

and thus & & # $ k(x2i − x2f ) (5.00 N/m) (0.100 m)2 − (0.080 m)2 vf = = = ±0.30 m/s. m 0.200 kg (7.43) where negative solution corresponds to the first time it reaches xf = 0.080 m. Example 7.9. A 2000 − kg (or 19600 − N) elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000−N frictional force to the elevator. What is the necessary force constant k for the spring.

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 88

Step 1: Choose a 1D coordinate system with x-axis pointing upwards and origin at the initial position (i.e. place of first contact of the elevator and the spring). Step 2: Initial conditions. At the initial state the kinetic energy is Ki =

1 (2000 kg) (4.00 m/s)2 = 16.0 kJ 2

(7.44)

the potential energy is 1 Ui = mgxi + kx2i = 0 2

(7.45)

Ei = Ui + Ki = 16.0 kJ.

(7.46)

and the total energy is

Step 3: Final conditions are given by Kf = 0 # $ # 1 1 Uf = mgxf + kx2f = (2000 kg) 9.80 m/s2 (−2.00 m) + k (−2.00 m)2 = −39.2 kJ + 2.00 m2 2# 2 $ Ef = −39 kJ + 2.00 m2 k (7.47)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 89 From the work energy theorem Wother = Ef − Ei # $ (17000 N) (−2.00 m) = −39.2 kJ + 2.00 m2 k − 16.0 kJ −34.0 kJ + 39.2 kJ + 16.0 kJ = 1.06 × 104 N/m. (7.48) k = 2.00 m2

7.3

Conservative and Nonconservative forces

We were able to define potential energy associated with work done by gravitational and elastic forces. All such forces are called conservative forces. Work done by conservative forces: • can always be expressed as difference between initial and final values of a suitably defined potential energy • it is reversible and is independent on the trajectory of the body, but only on initial and final points One might wonder if it is possible to do the same for all macroscopic forces which would allow to rewrite the work-energy theorem as a simple law of conservation of energy. It turns out that it is not possible and there are other (or non-conservative) forces for which it is not possible to define potential energy. For example, frictional force or air resistance forces are nonconservative. ⃗ = C x ˆj, Example 7.11. In a region of space the force of an electron is F where C is a positive constant. The electron moves around a square loop in ⃗ during the xy-plane. Calculate the work done on the electron by the force F a counterclockwise trip around square.

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 90

There are four legs of straight displacements around a closed path and so the total work is given by ' ( (L,0) ( (L,L) ( (0,L) ( ⃗ ⃗ W = F (x, y)·dl = Fx (x, 0)dx+ Fy (L, y)dy+ Fx (x, L)dx+ (0,0)

(L,0)

(L,L)

(7.49)

where (

(L,0)

Fx (x, 0)dx = 0 (0,0) (L,L)

(

Fy (L, y)dy = CL(L − 0) = CL2

(L,0)

(

(0,L)

Fx (x, L)dx = 0

(L,L)

(

(0,0)

Fy (0, y)dy = 0(0 − L) = 0

(7.50)

W = 0 + CL2 + 0 + 0 = CL2

(7.51)

(0,L)

and so which is non-zero and thus the force is non-conservative. The non-conservative forces cannot be described in terms of mechanical potential energies, but one can still associate with them other energies such as internal energy. For example the frictional force is non-conservative, but when friction is applied to objects in contact the internal properties of objects

(0,0)

Fy (0, y)dy (0,L)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 91 change. In particular friction leads to increase in temperature (or internal energy) which on the microscopic level corresponds to the change of kinetic energy of individual molecules. But from a macroscopic point of view one can think of work done by all non-conservative forces as a measure of change of internal energies Wother = −∆Uint (7.52) and the work energy theorem implies the conservation law of total energy ∆K + ∆U + ∆Uint = 0.

(7.53)

The big difference of the internal energy is that one cannot use it to do any useful work. This follows from the so-called second law of thermodynamics that we will see later in the course. The conservation law of energy and the second law of thermodynamics had passed a very large number of tests, but this does not stop people (usually outside of academia or theoretical physicists) from trying to build perpetual motion machines of the first kind (do work without input of energy) or of the second kind (do work using internal energy).

7.4

Force and potential energy

Let’s look more closely at work done on a particle when the displacement is small W = Fx (x)∆x (7.54) which can also be expressed as a change in potential energy W = −∆U.

(7.55)

By equating (7.54) and (7.55) we get Fx (x)∆x = −∆U

(7.56)

or in the limit of small infinitesimal displacement Fx (x) = − lim

∆x→0

For example

1 U = kx2 2

⇒

dU ∆U =− . ∆x dx

(7.57)

Fx (x) = −kx

(7.58)

Fx (x) = −mg.

(7.59)

or U = mgx

⇒

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 92 This can be easily generalized to 3D with dU dx dU = − dy dU = − dz

= − Fx = − lim∆x→0 ∆U ∆x Fy = − lim∆y→0

∆U ∆y

Fz = − lim∆z→0 ∆U ∆z

or in terms of unit vectors * ) ∂U(x, y, z)ˆ ∂U(x, y, z) ˆ ∂U(x, y, z) ˆ ⃗ i+ j+ k . F=− ∂x ∂y ∂z

(7.60)

(7.61)

It is convenient to define a vector-like operator also known as gradient (often pronounced as nabla) ) * ⃗ ≡ ∂ , ∂ , ∂ ∇ (7.62) ∂x ∂y ∂z or ˆ∂ ⃗ ≡ ˆi ∂ + ˆj ∂ + k (7.63) ∇ ∂x ∂y ∂z such that

⃗ = −∇U(x, ⃗ F y, z).

(7.64)

For example the gravitational potential energy is U = mgy

(7.65)

and so ⃗ = −∇(mgy) ⃗ F * ) ∂ ∂ ∂ ˆ ˆ ˆ +j +k (mgy) = − i ∂x ∂y ∂z ∂ (mgy) ˆ ∂ (mgy) ˆ ∂ (mgy) = −ˆi −j −k ∂x ∂y ∂z = −mgˆj.

(7.66)

Example 7.14. A puck with coordinates x and y slides on a level, frictionless air-hockey table. It is acted on by a conservative force described by the potential-energy function $ 1 # U(x, y) = k x2 + y 2 . 2

(7.67)

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 93 Find a vector expression for the force acting on the puck, and find an expression for the magnitude of the force. From (7.64) we get ) *) * $ 1 # 2 ˆ ∂ ⃗ = − ˆi ∂ + ˆj ∂ + k F k x + y2 ∂x ∂y ∂z 2 ˆ ˆ = −ikx − jky (7.68) and F =

7.5

+

(kx2 ) + (ky)2 = k

,

x2 + y 2 .

(7.69)

Energy Diagram

A convenient way of describing physical systems is in terms of energy diagrams plotting the potential energy and total energy as a function of displacement. For example consider a harmonic oscillator whose potential energy is 1 U = kx2 2

(7.70)

E = K + U.

(7.71)

and total energy is If there are no non-conservative forces, then the total energy can be represented by a constant line and potential energy by a parabola.

This is the so-called energy diagram whose main point is to study the possible solutions qualitatively without having to solve equations of motion quantitatively. Moreover, the energy graph can be used to identify stable (unstable) solutions corresponding to local minima (maxima) of the potential energy. For the example of harmonic oscillator a solution with x = 0 and dx =0 dt corresponds to a stable equilibrium solution. Then one can study small perturbations around the equilibrium which is what is done in field theories

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 94 where k plays the role of mass-squared parameter. However, if we imagine that the spring constant is negative k < 0, then the solution x = 0 and dx =0 is unstable. This phenomena is known as tachyonic instability which in dt the case in field theories corresponds to negative mass-squares (or imaginary masses in context of complex numbers). For more general potentials you can have both stable and unstable equilibriums. For example:

Which statement correctly describes what happens to the particle when it is at the maximum between x2 and x3 ? 1. The particle’s acceleration is zero. 2. The particle accelerates in the positive x-direction; the magnitude of the acceleration is less than at any other point between x2 and x3 . 3. The particle accelerates in the positive x-direction; the magnitude of the acceleration is greater than at any other point between x2 and x3 . 4. The particle accelerates in the negative x-direction; the magnitude of the acceleration is less at any other point between x2 and x3 . 5. The particle accelerates in the negative x-direction; the magnitude of the acceleration is greater than at any other point between x2 and x3 . The total energy was defined as a sum of kinetic and potential energies which

CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 95 are functions of position x and velocity )

dx E x, dt

*

1 = m 2

dx , dt

)

dx dt

e.g. *2

1 + kx2 . 2

(7.72)

but an even more fundamental quantity (called Lagrangian) is the difference between kinetic an potential energies ) * ) *2 dx 1 dx 1 L x, = m − kx.2 (7.73) dt 2 dt 2 Note that for any given trajectory x(t) the Lagrangian L is nothing but a function of time and thus can be integrated over some interval of time, i.e. * ( tf ) dx dt. (7.74) L x, dt ti Clearly, for any trajectory x(t) the integral in 7.74(also known as action functional) would be some real number and one might wonder what trajectory would produce the smallest (or largest) number? It turns out that such trajectories corresponding to stable (or unstable) classical solutions which is what you are most likely to observe in the lab. However, according to quantum mechanics many other solutions are also possible although their probabilities (or the so-called probability amplitudes) are suppressed exponentially, ! tf dx (7.75) ei ti L(x, dt )dt where the meaning of i will only be clear after you learn a (physicists version of) quantum mechanics.