Potential EnergyChapter 8

Energy ConservationAs mentioned last time, energy is an interesting quantity because it many cases it is conserved.

There are conservative and nonconservative forces. The easiest way to tell them apart is to ask if the work done in going from one place to another depends on the path taken.

If it does, then the force is nonconservative. But if the work done doesn’t depend on how you get from A to B this is a conservative force.

Conservative and Nonconservative Forces

Examples of conservative forces:

Gravity, Elastic force of a spring, Electric force

Examples of nonconservative forces:

Friction, air resistance, tension in a cord

One way to identify nonconservative forces is that the work done by them often goes into heat.

Potential EnergyWhen you throw a ball up into the air where does the energy go?

You do work on the ball with your arm, giving it velocity and thus kinetic energy, but as the ball gains height, its velocity decreases till it reaches zero. Where does this kinetic energy go?

The kinetic energy is turned into potential energy. Potential energy is the name for energy that is stored in an object because of the conditions it finds itself under.

If you drop a ball it accelerates down and gains kinetic energy. This energy comes from the gravitational potential energy it had as a result of you holding it above the ground.

Gravitational Potential EnergyLet’s consider the work done by gravity on an object that falls down a distance h.

We can define the change in potential energy as

Notice that gravitational potential energy is always the change between two heights. Since in principle something could always be lowered there is no end of the potential energy till you reach the center of the Earth.

W =

Z~F · d~l = Fd cos (0) = mgh

�U = U2 � U1 = mg(y2 � y1)

Elastic Potential EnergyNow lets consider the potential energy of a spring. Again we consider how much work is done when we stretch the spring by a distance x. However, now the force is not constant so

So the potential energy of the spring is just

W =

Z~F · dx =

Zkxdx =

1

2kx2

U(x) =1

2kx2

Example 8-3If the original height of rock is y1 = 3.0m, calculate the rocks speed when it has fallen to 1.o m above the ground.

1

2mv21 +mgy1 =

1

2mv22 +mgy2

gy1 =1

2v22 + gy2

=> v2 =p

2g(y1 � y2) =p2(9.80m/s2)(3.0m� 1.0m) = 6.3m/s

Example 8-7A dart of mass 0.100kg is pressed against the spring of a toy gun. The spring (with spring stiffness constant k=250N/m and ignorable mass) is compressed 6.0 cm and released. If the dart detaches from the spring when it reaches its natural length (x=0), what speed does the dart acquire?

knownm = 0.100kgk = 250N/mx = 0.060m

unknownv = ?

physics conservation of energy1

2mv21 +

1

2kx2

1 =1

2mv22 +

1

2kx2

2

0 +1

2kx2

1 =1

2mv22 + 0

=> v2 =kx2

m=> v =

s(250N/m)(0.060m)2

0.100kg= 3.0m/s

Potential Energy and ForceIn both examples we just did we found the potential energy by just integrating the force over distance. We can write a general expression for the potential energy associated with any force as:

Because of the relationship between integration and differentiation this means we can also write a force as the derivative of a potential:

U(x) = �Z

F (x)dx+ C

F (x) = �dU

dx

Gravitational PotentialWe have discussed the potential energy of gravity, but only near the Earth’s surface where the acceleration is close to constant. What about over large distances?

Then we should be able to get the potential energy by just integrating the force.

~F = GmM

r2r̂ U(r) = �

Z~F · d~r = �

ZGmM

r2dr = �GmM

Zdr

r2

U(r) = �GmM

r

where we have chosen the integration constant so that U(∞)= 0

Example 8-12A box of empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800m/s when 1600km above the Earth’s surface. The package eventually falls to the Earth. Estimate its speed just for impact.

knownv1 = 1800m/sr1 = 1600km = 1.6 × 106 m

unknownv2 = ?

1

2mv21 �G

mME

r1=

1

2mv22 �G

mME

r2

physicsconservation of energy

r2 = rE =6.38 × 106 mME = 5.98× 1024 kg

r1 = 7.98 × 106 m v2 =

rv21 � 2GME

� 1

r1� 1

r2

�

v2 =

s

(1800m/s)2 � 2(6.67⇥ 10�11Nm2

kg2)(5.98⇥ 1024kg)

� 1

7.98⇥ 106m� 1

6.38⇥ 106m

�

= 5320m/s

GoldenEye (1995)

What is the spring constant of the bungee cord?

This stunt was done for real and the dam used is in Switzerland. You can go today and make the jump yourself. The dam has a height of 220m. At the bottom before Bond shoots his grapple his velocity has slowed to zero. That means his original potential energy has now been entirely converted to the potential energy of the bungee cord which acts just like a spring.

where x is the distance the bungee cord has stretched. If we assume the bungee cord is 80m long then x = 220m – 80m = 140m where we have neglected the fact that Bond is a couple of meters from the bottom when he shoots his grapple. Note that the bungee cord is stretched to 2.75 times its original length, which is fairly typical for bungee cords. If we take Bonds mass to be 80kg, we then have

The upward force on Bond when he reaches the bottom is

mgh =1

2kx2

k =2mgh

x2=

2(80kg)(9.8m/s2)(220m)

(140m)2= 17.6N/m

F = kx = (17.6N/m)(140m) = 2, 464N

This force would give an acceleration of a = F/m = 2464N/80kg = 30.8m/s2

or 3.14g which is also typical in bungee jumping.

Power

Power is the rate at which work is done, or the rate at which energy is transformed.

The unit of power is the Watt, which is just energy over time or W= J/s.

In most practical situations we are concerned with power, since we want to use the energy is some time period.

P =dE

dt

PowerWe can write the power in terms of force and velocity.

P =dW

dtdW = ~F · d~l

for a force constant in time

P =d

dt(~F · d~l)

P = ~F · d~l

dt= ~F · ~v

P = ~F · ~v

Example 8-15Calculate the power required of a 1400kg car under the following circumstances. a) the car climbs a 10˚ hill (a fairly steep hill) at a steady 80 km/h. b) the car accelerates along a level road from 90 to 110km/h in 6.0s to pass another car. Assume that the average retarding force on the car is FR = 700N throughout.

mg

FR

Fc

a = 0

Fc = FR + mg sinθ = 700N + (1400kg)(9.8m/s2)(sin(10)) = 3100N

v = 80 km/h = 80 (1000m/1km)(1h/3600s) = 22m/s

P = ~F · ~v = 3100N · 22m/s = 6.8⇥ 104W

a)

b)a =

�v

�t=

110km/h� 90km/h

6s=

20km/h

6s= 0.93m/s2

XF = ma = F � FR => F = FR +ma = 700N + (1400kg)(0.93m/s2) = 2000N

P = ~F · ~v not constant, maximum for max v P = 2000N · 30.6m/s = 6.12⇥ 104W

Homework

Chapter 8: 24, 28, 32, 37, 39, 42, 69, 74