Motion, Forces and EnergyLecture 7: Potential Energy & Conservation

The name potential energy implies that the object in question has thecapability of either gaining kinetic energy or doing work on some other object:

Gravitational Potential Energy Ug = m g h

Elastic Potential Energy Us = ½ k x2

Conservation of Energy

h1

h2

0

y

v1

v2

Total Energy, E1

Total Energy, E2

022

1

2,2

2

1

2

1

2

1

22

21

21

22

21

2212

211

22212

gymmgyESo

gyvvthengyvvAs

vvmhhmgE

mvmghmvmghEEE

The change in total energy is E where:

Object in free fall

A ball of mass m is dropped from a height h above the ground.We can use energy calculations to find the speed of the ball whenit is either released from rest or with an initial speed vi:

h

y

yi = hUi = mghKEi = ½ mvi

2

yf = yUf = mgyKEf = ½ mvf

2

y=0Ug=0

)(2

:

)(2

)(2

2

1

2

1

:

2

22

22

yhgv

restFrom

yhgvv

yhgvv

mgymvmghmv

UKEUKE

Generally

f

if

if

fi

ffii

Energy losses (non-conservative forces)

Kinetic friction is an example of a non-conservative force. When a bookslides across a surface which is not frictionless, it will eventually stop.But all the KE of the book is NOT transferred to internal energy of the book.

We can find the speed of the mass sliding down the ramp below if we knowthe frictional force by considering kinetic and potential energies:

vi=0

vf

h=0.5m

d=1.0m

Initial energy, Ei = KEi+Ui = mgyi=14.7JFinal energy, Ef = KEf+Uf = ½ mvf

2

But Ei = Ef due to energy lossesdue to friction so energy loss E is

E = -fk.d = -k mgcos.d = -5.0J

So ½ mvf2 = 14.7-5.0 J

And therefore vf=2.54 ms-130o

Another similar problem:

The friction coefficient between the 3 kg block and the surface is 0.4 .If the system starts from rest, calculate the speed of the 5 kg ball when it has fallen a distance of 1.5 m.

Mechanical energy loss is –fk.d = -k m1g d (work done by friction on block) = -17.6 J

Change in PE for the 5 kg mass isU5kg = -m2 g d = -73.5 J

Without friction, the KE gained by BOTH masses would sum to U5kg (73.5 J), butWith friction, the KE gained = 73.5 – 17.6

= 55.9 J

KE = ½ (m1+m2)v2 (both masses move with same velocity)

So v = 3.7 ms-1.

m1=3.0 kg

m2=5.0 kg

fk

An inclined springA mass m starts from rest and slides a distance d down a frictionless incline.It strikes an unstressed spring (negligible mass) and slides a further distancex (compressing the spring which has a force constant, k). Find the initial separation of the mass and the end of the spring.

m

d

k

Vertical height travelled by the mass = (d+x)sin=h

Change in PE of the mass, Ug = 0 – mgh = -mg(d+x)sin

Elastic PE of spring increases from zero (unstressed) to:Us = ½ kx2

So Ug + Us = 0

So mg (d+x) sin = ½ kx2 giving d as:

xmg

kxd

sin2

2