PHY11 Lesson 1 Kinetic Energy, Work, And Power 2Q1415
-
Upload
ian-ag-adoctor -
Category
Documents
-
view
8 -
download
4
Transcript of PHY11 Lesson 1 Kinetic Energy, Work, And Power 2Q1415
Kinetic Energy, Work, and Power
PHY11 Lesson 1
Definition of Energy
• It is a (scalar) quantity that can be converted from one form to another but cannot be created nor destroyed.
• It is a quantity that can transferred from one body to another.
• It is the capacity of a body to do work.
Units of Energy
1 joule (J) = 1 N.m 1 ft-lb = 1.356 J
1 J = 107 ergs
1 J = 0.239 cal
1 Btu = 1055 J
1 eV = 1.602 x 10-19 J
Kinetic Energy (K)
• It is the energy associated with the state of motion of an object.
Conservation of Energy
• The total energy of a (closed) system does not change.
Consider an object of mass m moving with a speed v. The kinetic energy of the object is
21
2K mv
Work (W)
Work is energy transferred to or from an object by means of forces acting on or exerted by the object.
Consider a constant force F acting on an object of mass m that is free to move across a frictionless horizontal surface.
m
F
If the force causes the object to move through a displacement d, then the work done by the force on the object is
m
F
d
cosW Fd W F d ############################
Work and Kinetic Energy
Consider that, at t = 0, a constant force F acts on an object of mass m and initial velocity vo.
m F
vo
m F
d
v
From Newton’s 2nd Law: F ma Recall:
2 2
2ov v
ad
At time t > 0, the object would have a displacement d and a final velocity v.
2 2
2ov v
F md
2 21 1
2 2 oFd mv mv
Work-Kinetic Energy Theorem
2 21 1
2 2 oFd mv mv
work done
finalkineticenergy
initialkineticenergy
The total work done by a force (or forces) on an object is equal to the change in the kinetic energy of the object.
KKKW o
Problem 1
A bead with a mass 0.012 kg is moving along a wire in the positive direction of an axis. Beginning at time t=0, when the bead passes through x=0 with a speed 12 m/s, a constant force acts on the bead. The figure below indicates the bead’s position at these four times: to = 0, t1 = 1.0 s, t2 = 2.0 s, and t3 = 3.0 s. What is the kinetic energy of the bead at t = 10.0 s?
x(m)
0 5 10 15 20
to t1 t2 t3
Problem 1: Solution
The motion of the bead is UAM and its displacement is given by
From the figure, x = 10m when t1 = 1.0 s
22112)( atttx
221 )1()1(1210 a 2/0.4 sma
2212)( tttx tdt
dxv 412
At t = 10.0 s, smv /28)10(412
At t = 10.0 s, JsmkgK 7.4)/28)(012.0( 221
Problem 2
A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
Answers: (a) 2.4 m/s (b) 4.8 m/s
Problem 3
A 4.2 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x-axis along the track is applied to the body. The positions of the body as it slides to the right is shown in the figure. The force F is applied to the body at t = 0, and the graph records the position of the body at t = 1.00 s and t = 2.00 s. How much work is done on the body by the applied force between t = 0 and t = 1.5 s?
Answer: 0.756 J
x(m)
0 0.2 0.4
0.6
0.8
t=0
1.0s 2.0s
Work Done by Gravitational Force Fg
Consider an object of mass m thrown upward with an initial velocity vo and initial kinetic energy Ko =½ mvo
2.
vo
Fg
v
Fgd
As the object rises to a height d the work done by the gravitational force Fg is
180cosmgdW
mgdW
As an object rises, gravitational force does negative work on it and its kinetic energy decreases.
Work Done by Gravitational Force Fg
When the object has reached its maximum height, it will fall back down.
vo=0
Fgd
0cosmgdW
mgdW
As an object falls down, gravitational force does positive work on it and its kinetic energy increases.
Fg
v
The work done by gravitational force becomes
Work Done in Lifting/Lowering an Object
F
Fg
d
FdWa
As the object rises, the work done by the applied force is
Consider an object of mass m lifted by a force F.
F
Fg
The work done gravity is
mgdWg From the work-kinetic energy theorem,
gao WWKKK
Work Done in Lifting/Lowering an Object
F
Fg d
FdWa
If the object is being lowered, the work done by the force is
F
Fg
The work done gravity is
mgdWg
As an object is being raised, the work done by the applied force is positive while that of gravitational force is negative.
As an object is being lowered, the work done by the applied force is negative while that of gravitational force is positive.
Work Done by a Spring Force
Consider an ideal spring in its relaxed state (x=0).
where k is the called the spring constant or (force constant).
• The spring force is not constant.
• If Fx is (+) then x is (-) and vice versa.
kxFx
If a force is applied to stretch (or compress) the spring, the spring force is given by Hooke’s Law.
Work Done by a Spring Force
Consider an ideal spring (massless and obeys Hooke’s Law) of force constant k and a spring force Fx.
xFW aves
The spring force is a variable force and is directly proportional to x and the work done is
221 kxWs
2212
21 kxkxW os
xF
W xs
2
0
2221
os xxkW
Work Done by an Applied Force Consider a block of mass m attached to an ideal spring of force constant k.
m
Ki
m
x
Kf
FaFs
KKKWW osa
If there is no change in the kinetic energy of the block,
sa WW
Suppose a force Fa displaces the block along the x-axis. The work done by this applied force Wa plus the work done by the spring force Ws give
Work Done by a General Variable Force
Consider a variable force acting on an object in the +x-direction and that the force varies with the position x.
The work done by such variable force is
x
xodxxFW )(
Work Done by a Spring Force
)(2
1 22o
x
x
x
x xs xxkdxxkdxFWo
f
i
2212
21 kxkxW os
Work-Kinetic Energy Theorem with a Variable Force
Consider a variable force acting on an object in the +x-direction and that the force varies with the position x.
The work done by such variable force is
x
x
x
x
x
x ooo
dxdt
dvmdxmadxxFW )(
vdx
dv
dt
dx
dx
dv
dt
dv
v
v
v
v oo
dvvmdvmvW
KKKW o 22
2
1
2
1omvmvW
Work: Area Under the Curve
)( oxxFW
Constant Force
F(x)
xxo x
F
W
2
2
1
2
1
2
1kxxFbhW
Variable Force (F x)
F(x)
xx
W
F
General Variable Force
x
xodxxFW )(
xo x
Answers: (a) Fworker =74 N(b) Wworker =+333 J(c) Wf = -333 J(d) WN =0, Wg =0(e) WNet =0
Problem 4
Answers: (a) 30.3 m/s(b) 46.8 m
Problem 5
Answer: 2.90 m/s
Problem 6
the figure.
(see figure.)
Problem 7
Answers: (a) 0.600 m(b) 1.50 m/s
Problem 8
Answer: -209 J
Power
Power is the time rate at which work is done by a force.
t
WPavg
Average Power is the work done W by a force F in an amount of time t.
Instantaneous Power is the instantaneous time rate of doing work.
dt
dWP
Units of Power: slbftsJWwatt /.738.0/111
slbftWhphorsepower /.55074611
Power and Velocity
dt
dxF
dt
dWP
cos
cosFvP
vFP
Problem 9
Answer: 390 W
Problem 10
Answer: 28 passengers