1. Liceo Scientifico Isaac Newton
Physics course
Work
Professor
Serenella Iacino

2. Introduction
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.

3. 1 joule = 1 Newton ∙ 1 meter = 1kg ∙ s
Work → → F F θ θ →
Fcos θ s →
displacement s
fig.1
fig.2
W = F cos θ ∙ s
F ∙ s = F ∙ s ∙ cos θ → → m 2
1 joule = 1 Newton ∙ 1 meter = 1kg ∙ s 2

4. θ < 90° work is positive θ > 90° work is negative θ = 90°→ F
θ < 90° θ
fig.3 → s
work is positive → F
θ > 90° θ
fig.4 →
work is negative s → F
θ = 90° θ → s
work is zero
fig.5

5. When many forces act upon an object, we can obtain the total work by two equivalent methods:st
1 method:
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 = P ∙ s ∙ cos 90° = 0
W = N ∙ s ∙ cos 90° = 0
W = F ∙ s ∙ cos 180° = -F ∙ s
W = F ∙ s ∙ cos → P → → N F → N → θ F a → F a a a → → θ P F → s
W = -F s + F ∙ s ∙ cos θ
fig.6 ∙
sum a

6. W = F ∙ s ∙ cos 0° = F ∙ s = -F ∙ s + F ∙ s ∙cos θnd
2 method:
we can determine the resultant of all forces acting upon an object and then we determine the work of the resultant: → → s a y
F =-F
a = a → →
P = 0
N = 0
F =F cos θ x x x ax a x N F →
P =-P
N = N
F =F sen θ
F = 0
a = 0 F F θ y y y y ay y a ∑
F = m a x ∑ y
F = m a = 0 F x → P
fig.7 o x 2 2 2 F = ( ∑ F x ) + ( ∑ F y ) = ( ∑ F ) =
- F + F ∙ cos θ
W = F ∙ s ∙ cos 0° = F ∙ s = -F ∙ s + F ∙ s ∙cos
sum x a θ
sum
sum
sum a

7. 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. → → F → → F 1
W = F ∙ s = F ∙ s ∙ cos 0°= F ∙ s 1 1 1 1 o s s 1
fig.8

8. F
f(x) s Δ s s o 1 1
fig.9
W = ∫ f(x) dx s 1

9. Example 1 →
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: P h
fig.10 → →
W = P ∙ h = P ∙ h ∙ cos 0° = mgh

10. Example 2
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. →
W = P ∙ s = P ∙ s cos 180° = -mgh → →
ascent h
W = P ∙ s = P ∙ s cos 0° = + mgh → →
descent
fig.11

11. F = Kx , where K is the spring constant
The elastic force y F → el F x y
= F
= 0 → F → → F
= -Kx
elx F el F
= 0
ely x
fig.12
F = Kx , where K is the spring constant x
Hooke’s law is the relationship F
= - Kx
elx

12. Compressed Spring F = -F = -Kx → F → F F = 0 F → F = Kx → F F = 0 o yel F
= Kx →
elx F el F
= 0 o x
ely
fig.13

13. Work Graphic representation of the elastic force
y= Kx x x o s s o
y=-Kx
fig.14
fig.15
x ∙ (-Kx) -1 Kx 2
x ∙ (+Kx) +1 = Kx 2
W = =
W = 2 2 2 2

14. Kinetic Energy in Physics
v = v + a t v = a t
s = v t at s = a t f i
from which we have
then 1 2 1 2 i 2
W = F ∙ s = F s cos 0° = ma ∙ a t ∙ 1= m (a t) = m v → → 1 1 2 1 2 2 2 2 2
Kinetic Energy
( K ) is energy of motion ( measured in joules)

15. Variation of Kinetic Energy in Physics- v
t = f i a 1 v - v 1 v - v 2 v
s =
v t +
a t 2 = a f i + f i = i i 2 a 2 a 2 2 v - v v - v v - v v v v + - v v 1 = f i + a f i = f i f i = f i i a a a 2 2
2 a → → v 2 v 2 - - 1 2
m v f = i
W = F ∙ s = F s cos 0° = m a ∙ f i
2 a

16. The work - Energy Theorem
We observe that:
if the work is positive, then the kinetic energy increases when the object goes from the initial position to the final position;
if the work is negative, then the kinetic energy decreases;
if the work is zero, the kinetic energy doesn’t change, so the velocity is constant.

17. Example:
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. → →
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: → → a → N
W = W = 0
W = F ∙ s ∙ cos 180° = -F ∙ s = - 60,0 j
W = F ∙ s ∙ cos 0° = 200 j → → F F a → → N P → → P a a → F a s
fig.16 →
fig.17 F

18. we have: 1 1 W m v - m v = 2 2 1 - 60,0 + 200 = ∙ 10,0 ∙ ( v - 0 ) 2f i
sum 2 2 1
- 60, = ∙ 10,0 ∙ ( v ) 2 2 f
from which we have:
140
140
v = 2
v =
= 5,29 m s f 5 f 5

19. WORK
THE END