Introductory Physics

1. Work-Energy Theorem
Compiled by
Mphiriseni Khwanda

2. The Work-Energy Theorem and Kinetic Energy: horizontally along the plane
Consider a constant net external force acting on an object.
The object is displaced a distance s, in the same direction as
the net force.
The work is simply
s
F
   smasFW  
    2
2
12
2
122
2
1
ofof mvmvvvmasmW 
 axvv of 222
    22
2
1
of vvax 

3. When a net external force does work on and object, the kinetic
energy of the object changes according to
2
2
12
2
1
onet KEKEW omvmv 
THE WORK-ENERGY THEOREM: Horizontally

4. Work-Energy Theorem : vertically
 sFW cos
 
 
PE
hhmg
hhmgW o



0
gravity
hh fo
shence 
hh of
sbut 
Hence the work done by the force of gravity is equal to minus
the change in potential energy

5. Work-Energy Theorem: Combination of both Conservative and Nonconservative Forces
In normal situations both conservative and non-conservative forces act
simultaneously on an object, so the work done by the net external force can
be written as
nccnet
WWW 
KEKEKE of Wnet
PEPEPE fogravity  foc mghmghWW

6. ncc WWW 
ncW PEKE
THE WORK-ENERGY THEOREM becomes:
PEKE ncW
   ofof PEPEKEKEPEKE ncW
   ooff PEKEPEKE ncW
of EE ncW
If the net work on an object by nonconservative forces
is zero, then its energy does not change:
of EE 

7. 4.13 The Conservation of Mechanical Energy
THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY
The total mechanical energy (E = KE + PE) of an object
remains constant as the object moves, provided that the net
work done by external nonconservative forces is
In simple terms,
total mechanical energy before = total mechanical energy after
provided that the net work done by external nonconservative forces is
zero
What is the catch?
Condition for that to be true
zero.

8. 6.5 The Conservation of Mechanical Energy

9. SUMMARY: WORK-ENERGY THEOREM
Case 1: Horizontally
The work done by the net force (external, eg forces applied) on the object is equal to the change in the
object’s kinetic energy. 𝑊𝑛𝑒𝑡 = ∆𝐾 𝐸= (1
2
m𝑣2
− 1
2
m𝑣0
2
)
( 𝐹). 𝑑 = ∆𝐾 𝐸= (1
2
m𝑣2
− 1
2
m𝑣0
2
)
𝒎𝒂. 𝑑 = (1
2
m𝑣2
− 1
2
m𝑣0
2
)
Case 2: Vertically
The work done by the non-conservative forces (eg friction) on the object is equal the change in both the kinetic
and potential energies of the object
The work done by the net force(external, eg force of gravity) on the object is equal to minus the change in the
object’s potential energy
𝑊𝐶 = 𝑭 𝒈. 𝒅 = −∆𝑃𝐸= −mg(h − ℎ0)
𝒎𝒈. 𝒅 = −mg(h − ℎ0)
Case 3: The general case (the combination of the two)
𝑾 𝑵𝑪 = ∆𝑲 𝑬 + ∆𝑷 𝑬
𝑊𝑁𝐶 = ( 𝒇). 𝒅 = 𝝁𝒎𝒈. 𝒅 = 1
2m𝑣2
− 1
2m𝑣0
2
+ (mgℎ − 𝑚𝑔ℎ0)
𝑊𝑁𝐶 = 1
2m𝑣2
+ mgℎ − (1
2m𝑣0
2
+ 𝑚𝑔ℎ0) 𝑊𝑁𝐶 = 𝐸 − 𝐸0
Hence , the net work done by non-conservative forces is equal to the change in the energy of the system.
If 𝑊𝑁𝐶 = 0, hence 𝐸 = 𝐸0
1
2
m𝑣2
+ mgℎ = (1
2
m𝑣0
2
+ 𝑚𝑔ℎ0)

10. Problem solving strategy
Problem statement
Non-conservative forces present?
YES NO
𝑊𝑁𝐶 = 𝐸 − 𝐸0
𝑊𝑁𝐶 = 1
2
m𝑣2
+ mgℎ − (1
2
m𝑣0
2
+ 𝑚𝑔ℎ0)
1
2
m𝑣2
+ mgℎ = (1
2
m𝑣0
2
+ 𝑚𝑔ℎ0)
𝐸 = 𝐸0
Substitute and calculate the unknowns
( 𝒇). 𝒅 = 1
2m𝑣2
− 1
2m𝑣0
2
+ (mgℎ − 𝑚𝑔ℎ0)
Where Ʃ f represents sum of all non-conservative forces
present and d the distance covered during the application
of non-conservative forces

11. Applications: Work-energy theorem
1. A ramp in an amusement park is frictionless. A
smooth object slides down the ramp and
comes down through a height h, What
distance d is necessary to stop the object on
the flat track if the coefficient of friction is μ.
Section PQ is frictionless (keyword)
(Non-conservative forces absent)
Hence 𝑊𝑁𝐶 = 0,
𝐸 𝑄 = 𝐸 𝑃
1
2m𝑣 𝑄
2
+ mgℎ 𝑄 = 1
2m𝑣 𝑃
2
+ 𝑚𝑔ℎ 𝑃
1
2m𝑣 𝑄
2
+ mg(0) = 1
2m 0 2
+ 𝑚𝑔(ℎ)
1
2
m𝑣 𝑄
2
= 𝑚𝑔ℎ
Hence the total potential energy at p was
converted to kinetic energy at Q
Section QR has friction
Non-conservative forces present
Hence 𝑊𝑁𝐶 ≠ 0,
𝑊𝑁𝐶 = 𝐸 𝑅 − 𝐸 𝑄
( 𝒇). 𝒅 = 1
2
m𝑣 𝑅
2
+ mgℎ 𝑅 − (1
2
m𝑣 𝑄
2
+ 𝑚𝑔ℎ 𝑄)
𝝁𝒎𝒈. 𝑑 = 1
2m(0)2
+mgℎ 𝑅 − (𝑚𝑔ℎ + 𝑚𝑔ℎ 𝑄)
𝜇𝑚𝑔. 𝑑 = mgℎ 𝑅 − 𝑚𝑔ℎ − 𝑚𝑔ℎ 𝑄
Note that 𝒉 𝑹=𝒉 𝑸
𝜇𝑚𝑔. 𝑑 = −𝑚𝑔ℎ
𝒅 = −
𝒉
𝝁

12. Roller Coaster activity
Supposed a roller coaster starts from rest at point A and moves without friction as shown in the diagram.
(a)How fast is it moving at points B, C and D?
(b)What constant acceleration must be applied at D to have it stop at E?

13. Activity
The diagram on shows A 0.41-kg block sliding from A to B along a frictionless surface.
When the block reaches B, it continues to move along a horizontal surface BC where
a kinetic frictional force acts. As a result, the block slows down, coming to rest at C.
The kinetic energy of the block at A is 37J and heights above the ground.
(a)What is the kinetic energy of the block when it reaches B?
(b)How much work does the kinetic frictional force do during the BC segment trip?

14. Activity
A child of mass m is released from rest at the top of water slide, at a
height h = 9 m above the bottom of the slide as shown on the right.
Assuming that the slide is frictionless because of the water on it,
find the child’s speed at the bottom of the slide.