Physics 2514
Lecture 27
P. Gutierrez
Department of Physics & Astronomy
University of Oklahoma
Physics 2514 – p. 1/14
Goals
Introduce the concept of energy
Introduce kinetic energy;
Use gravity to introduce potential energy;
Arrive at the principle of energy conservation;
Discuss springs and their potential energy;
Discuss elastic collisions.
Physics 2514 – p. 2/14
Energy & Gravity
Consider a particle moving under the influence of gravity
m
dvy
= −mg
dt
Apply the chain rule from calculus
dy dvy
dvy
dvy
=
= vy
dt
dt dy
dy
⇒
dvy
= −mg
mvy
dy
Integrate the equation
Z
vf
mvy dvy = −
vi
Z
yf
mg dy
yi
⇒
1
1
2
mvf − mvi2 = −(mgyf − mgyi )
2
2
Physics 2514 – p. 3/14
Energy & Gravity
Integrate the equation
Z
vf
mvy dvy = −
vi
Z
yf
mg dy
yi
⇒
1
1
2
mvf − mvi2 = −(mgyf − mgyi )
2
2
Rearrange
1
1
2
mvf + mgyf = mvi2 + mgyi
2
2
Define 12 mv 2 as the kinetic energy
Define mgy as the gravitational potential energy
Physics 2514 – p. 4/14
Energy Conservation
Note kinetic energy plus potential energy initial and final are
equal (the sum of the two is a constant, the total mechanical
energy of the system E )
1
1
2
mvf + mgyf = mvi2 + mgyi = E
2
2
Solve for vf
vf2 = vi2 − 2g(yf − yi )
(Previously derived kinematic equation)
Solution depends only on change in position, therefore potential
energy is a relative quantity (can not give an absolute value to the
potential energy)
Physics 2514 – p. 5/14
Clicker
A 1 kg object is held 1 m above the floor, its total energy is:
A) 9.8 J
B) 0 J
C) 19.6 J
D) All of the above
Physics 2514 – p. 6/14
Summary
Kinetic energy K = 12 mv 2 energy due to motion, is always
positive
Gravitational potential energy U = mgy energy available to
convert to kinetic energy ∆K = −∆U
Total mechanical energy E = Ki + Ui is constant assuming
no frictional forces
Energy is a scalar (not a vector)
Zero of gravitational potential energy is arbitrary you decide
where to set it
Only potential energy differences matter
Energy units are kg-m2 /s2 = Joules
Physics 2514 – p. 7/14
Motion
Object thrown upward with initial velocity vy0 in the vertical
direction
Physics 2514 – p. 8/14
Energy Diagram
Total mechanical energy object in free fall: E = 12 mv 2 + mgy
Physics 2514 – p. 9/14
Energy Diagram
Total mechanical energy object in free fall: E = 12 mv 2 + mgy
Physics 2514 – p. 10/14
Clicker
A particle with the potential energy shown in the graph is moving
to the right. It has 1 J of kinetic energy at x = 1 m. Where is the
particle’s turning point?
A) x = 2 m
B) x = 3 m
C) x = 4 m
D) x = 5 m
E) x = 6 m
Physics 2514 – p. 11/14
Example Ballistic Pendulum
A m = 10 g bullet is fired into a M = 1200 g wood block that hangs
from a L = 150 cm long string. The bullet embeds itself into the block,
and the block swings out to an angle of θ = 40◦ . What was the speed
of the bullet?
Momentum conservation
mv0 = (M +m)v1
⇒
v0 =
M +m
v1
m
Energy conservation
v0 =
„
M +m
m
«
1
(m + M )v12 = (M + m)gy2
2
= (M + m)g(L − L cos θ)
p
⇒ v1 = 2gL(1 − cos θ) = 2.6 m/s
2.6 m/s = 317/ m/s
Physics 2514 – p. 12/14
Clicker
A small child slides down the four frictionless slides A-D. Each
has the same height. Her speeds at the bottom of the slide are
given by vA to vD . Which one of the following gives the correct
relation between speeds?
A) vD > vC > vA > vB ;
B) vD = vA > vB = vC ;
C) vA = vB = vC = vD
D) vA > vD > vB > vC
E) vA > vD = vB > vC
Physics 2514 – p. 13/14
Assignment
Start reading Chapter 10
Physics 2514 – p. 14/14