physics 111: mechanics lecture 6 wenda cao njit physics department
TRANSCRIPT
October 7-13, 2013
Chapter 6 Work and Kinetic Energy
6.1 Work 6.2 Kinetic Energy and the Work-Energy
Theorem 6.3 Work and Energy with Varying Forces 6.4 Power
October 7-13, 2013
Why Energy? Why do we need a concept of energy? The energy approach to describing
motion is particularly useful when Newton’s Laws are difficult or impossible to use.
Energy is a scalar quantity. It does not have a direction associated with it.
October 7-13, 2013
Kinetic Energy Kinetic Energy is energy associated with
the state of motion of an object For an object moving with a speed of v
SI unit: joule (J) 1 joule = 1 J = 1 kg m2/s2
2
2
1mvK
October 7-13, 2013
Work W
Start with Work “W”
Work provides a link between force and energy Work done on an object is transferred to/from it If W > 0, energy added: “transferred to the
object” If W < 0, energy taken away: “transferred from
the object”
xFmvmv x 20
2
2
1
2
1
October 7-13, 2013
Definition of Work W The work, W, done by a constant force on an
object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement
F is the magnitude of the force Δ x is the magnitude of the object’s displacement is the angle between
xFxFW
)cos(
and F x
October 7-13, 2013
Work Unit This gives no information about
the time it took for the displacement to occur the velocity or acceleration of the object
Work is a scalar quantity SI Unit
Newton • meter = Joule N • m = J J = kg • m2 / s2 = ( kg • m / s2 ) • m
xFxFW
)cos(
xFmvmv )cos(2
1
2
1 20
2
October 7-13, 2013
Work: + or -? Work can be positive, negative, or zero.
The sign of the work depends on the direction of the force relative to the displacement
Work positive: if 90°> > 0° Work negative: if 180°> > 90° Work zero: W = 0 if = 90° Work maximum if = 0° Work minimum if = 180°
sFsFW
)cos(
October 7-13, 2013
Example: When Work is Zero
A man carries a bucket of water horizontally at constant velocity.
The force does no work on the bucket
Displacement is horizontal Force is vertical cos 90° = 0
xFW )cos(
October 7-13, 2013
Example: Work Can Be Positive or Negative
Work is positive when lifting the box
Work would be negative if lowering the box The force would still be
upward, but the displacement would be downward
October 7-13, 2013
Work Done by a Constant Force
The work W done a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors: cosrFrFW
F
II
F
IIIr
F
I
r
F
IVr
r
0IW
cosrFWIV rFWIII
rFWII
October 7-13, 2013
Work and Force An Eskimo returning pulls a sled as shown.
The total mass of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30° and he pulls the sled 5.0 m ?
J
mN
xFW
2
2
102.5
)0.5)(30)(cos1020.1(
)cos(
October 7-13, 2013
Work Done by Multiple Forces
If more than one force acts on an object, then the total work is equal to the algebraic sum of the work done by the individual forces
Remember work is a scalar, so this is the algebraic sum
net by individual forcesW W
rFWWWW FNgnet )cos(
October 7-13, 2013
Kinetic Energy Kinetic energy associated with the
motion of an object Scalar quantity with the same unit as
work Work is related to kinetic energy
2
2
1mvK
xFmvmv net 20
2
2
1
2
1
KKKW ifnet
October 7-13, 2013
Work-Energy Theorem When work is done by a net force on an
object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy
Speed will increase if work is positive Speed will decrease if work is negative
20
2
2
1
2
1mvmvWnet
KKKW ifnet
October 7-13, 2013
Problem Solving Strategy Identify the initial and final positions of the body,
and draw a free body diagram showing and labeling all the forces acting on the body
Choose a coordinate system List the unknown and known quantities, and
decide which unknowns are your target variables Calculate the work done by each force. Be sure to
check signs. Add the amounts of work done by each force to find the net (total) work Wnet
Check whether your answer makes sense
October 7-13, 2013
Work and Kinetic Energy The driver of a 1.00103 kg car traveling on the
interstate at 35.0 m/s slam on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the breaks are applied, a constant friction force of 8.00103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collisions occur?
October 7-13, 2013
Work and Kinetic Energy (a) We know Find the minimum necessary stopping distance
Nfkgmvsmv k33
0 1000.8,1000.1,0,/0.35
233 )/0.35)(1000.1(2
1)1000.8( smkgxN
22
2
1
2
1iffricNgfricnet mvmvWWWWW
202
10 mvxfk
mx 6.76
October 7-13, 2013
Work and Kinetic Energy (b) We know Find the speed at impact. Write down the work-energy theorem:
Nfkgmvsmv k33
0 1000.8,1000.1,0,/0.35
Nfkgmsmvmx k33
0 1000.8,1000.1,/0.35,0.30
xfm
vv kf 22
02
22
2
1
2
1ifkfricnet mvmvxfWW
2233
22 /745)30)(1000.8)(1000.1
2()/35( smmN
kgsmv f
smv f /3.27
October 7-13, 2013
Work with Varying Forces On a graph of force as a function
of position, the total work done by the force is represented by the area under the curve between the initial and the final position
Straight-line motion
Motion along a curve
...... bbxaax xFxFW
dxFWx
x x 2
1
ldFdlFdlFWP
P
P
P
P
P
2
1
2
1
2
1
cos ||
October 7-13, 2013
Work-Energy with Varying Forces
Work-energy theorem Wtol = K holds for varying forces as well as for constant ones
dx
dvv
dt
dx
dx
dv
dt
dva x
xxx
x
dxdx
dvmvdxmadxFW
x
x
xx
x
x x
x
x xtot 2
1
2
1
2
1
KmvmvWtot 21
22 2
1
2
1
x
v
v xtot dvmvW 2
1
October 7-13, 2013
Spring Force: a Varying Force
Involves the spring constant, k
Hooke’s Law gives the force
F is in the opposite direction of x, always back towards the equilibrium point.
k depends on how the spring was formed, the material it is made from, thickness of the wire, etc. Unit: N/m.
dkF
October 7-13, 2013
Measuring Spring Constant Start with spring at its
natural equilibrium length. Hang a mass on spring
and let it hang to distance d (stationary)
From
so can get spring constant.
April 19, 2023
0xF kx mg
mgk
d
October 7-13, 2013
Work done on a Spring To stretch a spring, we must do
work We apply equal and opposite
forces to the ends of the spring and gradually increase the forces
The work we must do to stretch the spring from x1 to x2
Work done on a spring is not equal to work done by a spring
21
22 2
1
2
12
1
2
1
kxkxdxkxdxFWx
x
x
x x
October 7-13, 2013
Power Work does not depend on time interval The rate at which energy is transferred is
important in the design and use of practical device
The time rate of energy transfer is called power
The average power is given by
when the method of energy transfer is work
WP
t
October 7-13, 2013
Instantaneous Power Power is the time rate of energy transfer.
Power is valid for any means of energy transfer
Other expression
A more general definition of instantaneous power
vFt
xF
t
WP
vFdt
rdF
dt
dW
t
WP
t
0lim
cosFvvFP
October 7-13, 2013
Units of PowerThe SI unit of power is called the
watt 1 watt = 1 joule / second = 1 kg . m2 / s3
A unit of power in the US Customary system is horsepower 1 hp = 550 ft . lb/s = 746 W
Units of power can also be used to express units of work or energy 1 kWh = (1000 W)(3600 s) = 3.6 x106 J
October 7-13, 2013
A 1000-kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?
Power Delivered by an Elevator Motor
yynet maF ,
0 MgfT
NMgfT 41016.2
W
smNFvP4
4
1048.6
)/3)(1016.2(
hpkWP 9.868.64