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