Post on 27-Mar-2015
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