Liceo Scientifico Isaac Newton

Physics course

Work

Professor

Serenella Iacino

First of all in Physics work is always done by forces, in fact a force acting upon an object does work when the initial point of application of the force is displaced.

Introduction

Work

F

Fcos

displacement s

θ

W = F cos θ ∙ s

fig.1

θ

→

→

F ∙ s = F ∙ s ∙ cos θ→ →

fig.2

F

s

θ

→

→

sm 2

2 1 joule = 1 Newton ∙ 1 meter = 1kg ∙

s

fig.3

θ F

→

→

work is positive

s

fig.4

F

→

→

work is negative

fig.5 work is zero

→

θ < 90°

θ

s

F

→θ

θ > 90°

θ = 90°

When many forces act upon an object, we can obtain the total work by two equivalent methods:

P

F

F

N→ →

→

→a

θ

fig.6

P→

N→

a→

aF a

θF→

s→

W = P ∙ s ∙ cos 90° = 0

W = N ∙ s ∙ cos 90° = 0

W = F ∙ s ∙ cos 180° = -F ∙ s

W = F ∙ s ∙ cos

1 method:st

we can determine the work done by each force acting upon an object and then we add them to obtain the sum of work:

W = -F s + F ∙ s ∙ cos sum a∙ θ

P

F

F

→

→a

x

y→

N→

θ

fig.7

Fx

Fy

s→

a→

∑F = m ax x

P = 0x

y yP =-P

N = 0

N = N

x

y

F =F cos

F =F sen

θ

θ

xF =-Fax

F = 0ay

a a = a

a = 0y

x

o

2 method:nd

we can determine the resultant of all forces acting upon an object and then we determine the work of the resultant:

- - F + F ∙ cos sum

∑ Fx

∑ Fy

22

∑x

2

aθF )( + ( ) = F( ) ==

∑yyF = m a = 0

W = F ∙ s ∙ cos 0° = F ∙ s = -F ∙ s + F ∙ s ∙cos sum sum sum a θ

We can represent graphically on a Cartesian plane the work of a constant force F that is parallel to the displacement s and has the same direction, placing the displacement along the x axis and the force along the y axis.

s

F

o

F1

1s

fig.8

W = F ∙ s = F ∙ s ∙ cos 0°= F ∙ s 11

→ →1 1

→ →

F

so

fig.9

s1

W = ∫ f(x) dx1s

0

f(x)

Δs1

If an object, having mass m, falls from a height h, the angle between the gravity force and the displacement is zero, so the work done by this force is:

→→

fig.10

h

Example 1

P→

W = P ∙ h = P ∙ h ∙ cos 0° = mgh

During a football match, a ball of mass m is kicked up, reaching a height h and then falls to the ground. Calculate the work done by the gravity force P during the ascent and during the descent.

→

ascent

→ →

descent

→ →

fig.11

h

W = P ∙ s = P ∙ s cos 180° = -mgh

W = P ∙ s = P ∙ s cos 0° = + mgh

Example 2

x

y

fig.12

FF

F

x

y

= F

= 0

→

F→

el

F→

The elastic force

xF = Kx , where K is the spring constant

F→ F

F

elx= -Kx

= 0el

ely

Felx = - KxHooke’s law is the relationship

Compressed Spring

FF

F

x

y

= -F = -Kx

= 0

→

FF

F

elx

ely

= Kx

= 0

→el

x

y

o

fig.13

F→

F→

el

F

o

y=-Kx

x

fig.14

Work Graphic representation of the elastic force

x ∙ (-Kx)2 2

W = = -1 2Kx

s

y= Kx

x

fig.15

F

o s

2W = x ∙ (+Kx) =

2+1 2Kx

Kinetic Energy in Physics

2from which we have thenif

i 2

21

v = v + a t v = a t

s = v t + at s = a t2

1

W = F ∙ s = F s cos 0° = ma ∙ a t ∙ 1= m (a t) = m v →→ 2221

2

1

2

1

2

Kinetic Energy ( K ) is energy of motion ( measured in joules)

2

t =

2i

f i

a

v v

f i

a

v v

2f i

a

v v+

--

-

2

f i

a

v v-=

11

iv a+

21 f i

a

v v- f i

2

v v+=

=s =

=

avv t + i a t =

f i

a

v v-

f i

2 a

v v-22

W = F ∙ s = F s cos 0° = m a ∙ →→

f i

2 a

v v-22

-1

2m v

f2

= 1

2m v

i2

Variation of Kinetic Energy in Physics

We observe that:

1. if the work is positive, then the kinetic energy increases when the object goes from the initial position to the final position;

2. if the work is negative, then the kinetic energy decreases;

3. if the work is zero, the kinetic energy doesn’t change, so the velocity is constant.

The work - Energy Theorem

a man pushes a supermarket trolley of mass m = 10 kg, for two metres, applying a force F the magnitude of which is 100 N; the friction force between the floor and the trolley is 30 N.

P

FF

N→

→

→

→a

fig.17

s→

W = W = 0

W = F ∙ s ∙ cos 180° = -F ∙ s = - 60,0 j

W = F ∙ s ∙ cos 0° = 200 j F

→

F→

P→

N→

a

a a

→

→→

→a

Knowing that the force of gravity P and the normal force N also act upon the object, we draw the free-body diagram and we determine the work done by each force:

Example:

fig.16

we have:

sum-1

2m v

f2

=1

2m v

i2W

1

2- 60,0 + 200 = ∙ 10,0 ∙ ( v - 0 ) 2

f

from which we have:

v =f

2 1405

1405

v =f

= 5,29 m s

THE ENDWORK