Transcript Motion, Forces and Energy Lecture 7: Potential Energy & Conservation

Motion, Forces and Energy
Lecture 7: Potential Energy & Conservation
The name potential energy implies that the object in question has the
capability 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
v1
Total Energy, E1
v2
Total Energy, E2
y
h2
0
The change in total energy is E where:
1
1


2
2
E  E2  E1   mgh2  mv2    mgh1  mv1 
2
2

 

1
2
2
E  mg h2  h1   m v2  v1
2
2
2
2
2
As v2  v1  2 gy, then v1  v2  2 gy

1
So E  mgy  m  2 gy   0
2

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 when
it is either released from rest or with an initial speed vi:
Generally :
KEi  U i  KE f  U f
yi = h
Ui = mgh
KEi = ½ mvi2
yf = y
Uf = mgy
KEf = ½ mvf2
h
y
1
1
2
2
mvi  mgh  mv f  mgy
2
2
2
2
v f  vi  2 g (h  y )
v f  vi  2 g (h  y )
2
From rest :
v f  2 g (h  y )
y=0
Ug=0
Energy losses (non-conservative forces)
Kinetic friction is an example of a non-conservative force. When a book
slides 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 know
the frictional force by considering kinetic and potential energies:
Initial energy, Ei = KEi+Ui = mgyi=14.7J
Final energy, Ef = KEf+Uf = ½ mvf2
vi=0
d=1.0m
But Ei = Ef due to energy losses
due to friction so energy loss E is
E = -fk.d = -mk mgcosq.d
= -5.0J
h=0.5m
So ½ mvf2 = 14.7-5.0
30o
vf
J
And therefore vf=2.54 ms-1
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 = -mk m1g d
(work done by friction on block) = -17.6 J
m1=3.0 kg
Change in PE for the 5 kg mass is
U5kg = -m2 g d = -73.5 J
fk
Without friction, the KE gained by BOTH
masses would sum to U5kg (73.5 J), but
With friction, the KE gained = 73.5 – 17.6
= 55.9 J
m2=5.0 kg
KE = ½ (m1+m2)v2 (both masses move
with same velocity)
So v = 3.7 ms-1.
An inclined spring
A 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 distance
x (compressing the spring which has a force constant, k). Find the initial
separation of the mass and the end of the spring.
m
d
Vertical height travelled by the mass = (d+x)sinq=h
Change in PE of the mass, Ug = 0 – mgh = -mg(d+x)sinq
Elastic PE of spring increases from zero (unstressed) to:
Us = ½ kx2
k
q
So Ug + Us = 0
So mg (d+x) sinq = ½ kx2 giving d as:
kx2
d
x
2mg sin q