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Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Lecture 14
Chapter 9
Spring Force and Power
Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Today we are going to discuss:
Chapter 9:
Scalar (Dot) product of two vectors: Section 9.3 Work done by a Spring: Section 9.4 Skip: Section 9.5 Power: Section 9.6
IN THIS CHAPTER, you will learn how to solve problems using two new concepts: work and kinetic energy instead of forces.
.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
In order to simplify the work expression, let’s introduce very useful
Dot Product of Vectorsor
Scalar product
Let’s digress from work for a few slides and then, we will get “back to work”
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Scalar(Dot) Product of Two VectorsThe scalar product of two vectors is:
Therefore, we can say that work is a scalar product of force and displacement
Workdone by F
∙Workdone by F
A
B kAjAiAA zyx
ˆˆˆ
kBjBiBB zyxˆˆˆ
In a component form:
zzyyxx BABABABA
In a magnitude/angle form:
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
A constant force F acts on an rabbit as it moves from position r1 to r2. What is the work done by this Force?
kjir ˆ3ˆ4ˆ21 kjir ˆ6ˆ3ˆ52
kjiF ˆ2ˆ5ˆ4
rs
JsFW 29)9)(2()7)(5()3)(4(
12 rr kji ˆ9ˆ7ˆ3
Example
F
1r
2r
s
∙Workdone by F
In a baseball game, the catcher
stops a 90-mph pitch. What can
you say about the work done by
the catcher on the ball?
A) catcher has done positive work
B) catcher has done negative work
C) catcher has done zero work
The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F s cos ), because = 180º, then W < 0. Note that because the work done on the ball is negative, its speed decreases.
ConcepTest Play Ball!
F
s 0W
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.
Problem 9.33Example
The work W0 accelerates a car from 0 to v0. How much work is needed to accelerate the car from v0 to 3v0?
A) 2 W0
B) 3 W0
C) 6 W0
D) 8 W0
E) 9 W0
ConcepTest Speeding Up
KWnet Work-Kinetic Energy Principle
20
2200 2
02
:1 vmvmKWCase
020
20
20 8
28)3(
2:2 WvmvvmKWCase
ConcepTest Tension and Work
A) tension does no work at all
B) tension does negative work
C) tension does positive work
A ball tied to a string is being whirled around in a circle. What can you say about the work done by tension?
v T
No work is done because the force
acts in a perpendicular direction to
the displacement. Or using the
definition of work (W = F s cos ),
because = 90º, then W = 0.
Follow-up: Does the Earth do work on the Moon?
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Work Done By a Varying Force
We need to figure out how to deal with these cases.
kxF
In the previous class, we were finding work done by constant forces
Object oscillating on a spring
But there are forces which are not constant, let’s look at one of those, spring force
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
The Spring Force – variable forceThe variable force exerted by a spring is given by Hooke’s Law:
kxFspring
0xx
xx
0sF0x0 kxFs
0x0)( xkFs
0x
The force is to the left
It is called a Restoring force
The force is to the right
SF
k – spring constantEquilibrium Equilibrium Equilibrium
SF
0x
The spring force returns the cart to the equilibrium.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Work done by a spring
1
1
nxdxx
nn
Spring force:
dx - displacement
Work done by Fsp:This is used to integrate:
x0x dxspF
StretchedkxFspring
22
2 ifsp xxkW
Equilibrium
fxix
dxFWf
i
x
xspsp dxkx
f
i
x
x )( dxxk
f
i
x
x
f
i
x
x
xk
2
2
Let’s calculate work done a spring force
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Work is an area under a curve F-vs-x
;111 xFW
W W1 W2 W3 ....
x
etcxFW ;222
iix
xFWi 0
lim
Total work done by F is a sum:
7
1
i
iiWW
By definition, this is an integral:
Work done by F over each ∆x:
∆ ∆ ∆
∆∆
∆∆
dxFWf
i
x
xx
Graphical meaning of an integral is an area under a curve F-vs-x.Let’s convince ourselves:
7
1
i
iii xF
We must evaluate the integral either geometrically, by finding the area under the curve, or by actually doing the integration.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Power
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Power
Units Watts = Joules/sec
dtdWP
The instantaneous Power is the rate at which work is done
dtsdF
dtsdF
vF
The average Power is the work done divided by the time it takes to do the work.
workthisdototakentimeforceabydonework
tWP
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
A certain 1000 kg car can accelerate from rest to a speed of 20 m/s in a time of 10 s. What average power must the motor produce in order to cause this acceleration?
00 v smv f /20
The work done by the motor in accelerating the car can be found from the work-KE principle:
workthisdototakentimeforceabydonework
tWP By definition, the average Power is
KW if KK 0
2
2fmv
10 s is the time taken for this work
tWP
tmv f
221
kWWatts
smkg 202000010
)/20(1000 221
Average Car PowerExample
Mike performed 5 J of work in 10 secs. Joe did 3 J of work in 5 secs. Who produced the greater power?
A) Mike produced more power
B) Joe produced more power
C) both produced the same amount of power
Because power = work / time, we see that Mike produced 0.5 W and Joe produced 0.6 W of power. Thus, even though Mike did more work, he required twice the time to do the work, and therefore his power output was lower.
ConcepTest Time for Work
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
How much power does it take a 50-kg runner to run up a 5 m high hill in 10 s? Assume acceleration is zero.
workthisdototakentimeforceabydoneworkP
By definition, the average Power is
Average Runner PowerExample
Department of Physics and Applied PhysicsPHYS.1410 Lecture 14 Danylov
Thank youSee you on Wednesday