2. Work and Energy
Principle of Conservation of Energy
“Energy can neither be created nor
destroyed in any process. It can be
converted from one form to another or
transferred from one body to another,
but the total amount remains constant”
Work
;the product of a force times the
distance through which it acts
𝑊 = 𝐹𝑠
***𝐹 and 𝑠 are both vectors, but 𝑊 is
scalar
Multiplication of vectors
; Dot product (scalar)
; Cross product (vector)
Dot product
Suppose
𝐴 = 𝑎 𝑥 𝑖 + 𝑎 𝑦 𝑗 + 𝑎 𝑧 𝑘
𝐵 = 𝑏 𝑥 𝑖 + 𝑏 𝑦 𝑗 + 𝑏 𝑧 𝑘
𝐴 ∙ 𝐵 = 𝑎 𝑥 𝑏 𝑥 + 𝑎 𝑦 𝑏 𝑦 + 𝑎 𝑧 𝑏 𝑧
𝐴 ∙ 𝐵 = 𝐴 𝐵 𝑐𝑜𝑠𝜃
𝐴
𝐵
𝜃
𝐴 ∙ 𝐵 = 𝐴 𝐵 𝑐𝑜𝑠𝜃
𝐹
𝜃 𝐹 𝑐𝑜𝑠𝜃
𝐹 𝑠𝑖𝑛𝜃
∆𝑟
From the diagram,
𝑊 = 𝐹∆𝑟𝑐𝑜𝑠𝜃
𝑊 = 𝐹 ∆𝑟 𝑐𝑜𝑠𝜃
𝑊 = 𝐹 ∙ ∆𝑟
*** 𝑊 = 𝐹 ∙ ∆𝑟 is more accurate
***In reality, 𝐹 is not applied constantly
and ∆𝑟 is not always a straight line
Concept
***For a curve path, see the curve as the
combination of increadibly tiny straight
lines, 𝑑𝑟
4 types of Work
i. 𝐹; constant magnitude and direction
ii. 𝐹; inconstant magnitude and constant
direction
iii. 𝐹; constant magnitude and inconstant
direction
iv. 𝐹; inconstant magnitude and direction
𝑊 = 𝐹 ∙ ∆𝑟
𝑑𝑊 = 𝐹 ∙ 𝑑𝑟
𝑊𝑡𝑜𝑡𝑎𝑙 = 𝑑𝑊 = 𝐹 ∙ 𝑑𝑟
3. Uniform Force
;both magnitude and direction are
constant
𝑊𝐴→𝐵 = 𝑑𝑊
𝐵
𝐴
= 𝐹 ∙
𝐵
𝐴
𝑑𝑟
Since, 𝐹 is uniform,
𝑊𝐴→𝐵 = 𝐹 ∙ 𝑑𝑟
𝐵
𝐴
𝑊𝐴→𝐵 = 𝐹 ∙ (𝑟𝐵 − 𝑟𝐴)
𝑊𝐴→𝐵 = 𝐹 ∙ ∆𝑟𝐴→𝐵
***𝑊 is independent on the path, but is
depend on the displacement when 𝐹 is
uniform. A uniform force is considered
“Conservative Force”
𝑊 > 0; work done by the system
𝑊 < 0; work done on the system
𝑟𝐴
𝑥
𝑦
𝑜
𝑟𝐵
∆𝑟
𝐵
𝐴
𝑊𝐴→𝐵 = 𝐹 ∙ ∆𝑟𝐴→𝐵
Non-uniform Force
;inconstant magnitude and constant
direction
Ex; Spring
At (a), x=0, equilibrium point
At (b), an external force (𝐹𝑒𝑥𝑡) pulls an
object to the right, restoring force (𝐹𝑠)
will be experienced by the object also
∴ 𝐹𝑒𝑥𝑡 = 𝐹𝑠
At (b),𝐹𝑒𝑥𝑡 = 𝑘𝑥1
At (c),𝐹𝑒𝑥𝑡 = 𝑘𝑥2
𝑑𝑊 = 𝐹 ∙ 𝑑𝑟
𝑊 𝑏 →(𝑐) = 𝑑𝑊
(𝑐)
(𝑏)
= 𝐹𝑒𝑥𝑡 ∙
(𝑐)
𝑏
𝑑𝑟
= 𝐹𝑒𝑥𝑡
(𝑐)
𝑏
𝑑𝑟 𝑐𝑜𝑠𝜃
= 𝑘𝑥
𝑐
𝑏
𝑑𝑥 (𝑐𝑜𝑠0°)
m
m
m
(𝑐)
(𝑏)
(𝑎)
𝑥1
𝑥 = 0
𝐹𝑒𝑥𝑡
𝐹𝑠
𝐹𝑠 = 𝑟𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
𝑥2
4. 𝑊 𝑏 →(𝑐) = 𝑘𝑥
𝑐
𝑏
𝑑𝑥 (𝑐𝑜𝑠0°)
𝑊 𝑏 →(𝑐) =
1
2
𝑘𝑥2
2
−
1
2
𝑘𝑥1
2
Since,
1
2
𝑘𝑥2
is Elastic Potential Energy
*** 𝑊𝐴→𝐵 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝐹𝑒𝑥𝑡
***𝑊 𝐹𝑠
= 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦𝐹𝑠
Non-uniform Force
;constant magnitude and inconstant
direction
Ex; Circular motion
𝑑𝑊 = 𝐹 ∙ 𝑑𝑟 = 𝐹 𝑑𝑟 𝑐𝑜𝑠𝜃
Since at every point, 𝜃 = 90°
𝑊 = 0
𝑊𝐴→𝐵 = 𝐸𝑃𝐸 𝐵 − 𝐸𝑃𝐸𝐴
𝑊 𝐹𝑒𝑥𝑡
= −𝑊 𝐹𝑠
𝑣0
𝑣0
𝐹𝐴
𝐹𝐵
𝐴
𝐵
Non-uniform Force
;inconstant magnitude and direction
***non-conservative force; work done by
force depends on the paths taken by a
particle
Ex; a particle moves from A to B in AEB
and ADCB paths. The equation of force
in the field is 𝐹 = −𝑦𝑖 + 𝑥𝑗 Newton.
Find the work done in the 2 paths.
The external force acting on the particle
need to be equal to the force of the field
and has opposite direction to cancel out.
∴ 𝐹𝑒𝑥𝑡 = −𝐹
𝑊𝐴→𝐸→𝐵 = 𝐹𝑒𝑥𝑡 ∙ 𝑑𝑟
𝐵
𝐴
= −𝐹 ∙ 𝑑𝑟
𝐵
𝐴
= −
−𝑦
𝑥
∙
𝑑𝑥
𝑑𝑦
𝐵
𝐴
= 𝑦𝑑𝑥
𝐵
𝐴
− 𝑥𝑑𝑦
𝐵
𝐴
= 𝑦𝑑𝑥
𝑥=1
𝑥=1
− 1 𝑑𝑦
𝑦=−1
𝑦=1
= 0 − −1 − 1 = 2 𝐽𝑜𝑢𝑙𝑒𝑠
𝑥
𝑦
𝐴(1,1) 𝐷(2,1)
𝐵(1, −1) 𝐶(2, −1)
𝐸
0
5. 𝑊𝐴→𝐷→𝐶→𝐵 = 𝑊𝐴→𝐷 + 𝑊𝐷→𝐶 + 𝑊𝐶→𝐵
= 𝑦𝑑𝑥
𝐷
𝐴
− 𝑥𝑑𝑦
𝐷
𝐴
+ 𝑦𝑑𝑥
𝐶
𝐷
− 𝑥𝑑𝑦
𝐶
𝐷
+ 𝑦𝑑𝑥
𝐵
𝐶
− 𝑥𝑑𝑦
𝐵
𝐶
= (1)𝑑𝑥
𝑥=2
𝑥=1
− 𝑥𝑑𝑦
𝑦=1
𝑦=1
+ 𝑦𝑑𝑥
𝑥=2
𝑥=2
− (2)𝑑𝑦
𝑦=−1
𝑦=1
+ (−1)𝑑𝑥
𝑥=1
𝑥=2
− 𝑥𝑑𝑦
𝑦=−1
𝑦=−1
= (1)𝑑𝑥
𝑥=2
𝑥=1
− 0 + 0 − (2)𝑑𝑦
𝑦=−1
𝑦=1
+ (−1)𝑑𝑥
𝑥=1
𝑥=2
− 0
= (1)𝑑𝑥
𝑥=2
𝑥=1
− 2 𝑑𝑦
𝑦=−1
𝑦=1
+ (−1)𝑑𝑥
𝑥=1
𝑥=2
= 2 − 1 − −2 − 2 + −1 − −2
= 1 + 4 + 1 = 6 Joules
𝑥
𝑦
𝐴(1,1) 𝐷(2,1)
𝐵(1, −1) 𝐶(2, −1)
𝐸
0
Prove that 𝐹 = 2𝑥𝑖 + 3𝑦2
𝑗 is a
conservative force.
Let the initial and final point be 𝐴 and 𝐵
respectively
𝑊𝐴→𝐵 = 𝐹 ∙ 𝑑𝑟
𝐵
𝐴
=
2𝑥
3𝑦2 ∙
𝑑𝑥
𝑑𝑦
𝐵
𝐴
= 2𝑥𝑑𝑥
𝐵
𝐴
+ 3𝑦2
𝑑𝑦
𝐵
𝐴
= 𝑥2
+ 𝑦3
𝐴
𝐵
Since, the value of 𝑊𝐴→𝐵 depends on
only the values of 𝑥 and 𝑦 at 𝐴 and 𝐵. It
does not depend on the other points on
the path. Thus, 𝐹 is a conservative force.
***How to check whether 𝑭 is a
conservative force?
For
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘
𝐹 is a conservative force, when
𝜕𝐹𝑥
𝜕𝑦
=
𝜕𝐹𝑦
𝜕𝑥
And
𝜕𝐹𝑦
𝜕𝑧
=
𝜕𝐹𝑧
𝜕𝑦
And
𝜕𝐹𝑧
𝜕𝑥
=
𝜕𝐹𝑥
𝜕𝑧
Power
; average and instantaneous power
𝑃 =
𝑊
∆𝑡
𝑃 =
𝑑𝑊
𝑑𝑡
6. 𝑃 =
𝑑𝑊
𝑑𝑡
𝑃 =
𝑑𝑊
𝑑𝑡
Since, 𝑊 = 𝐹 ∙ 𝑑𝑟
𝑃 =
𝑑( 𝐹 ∙ 𝑑𝑟)
𝑑𝑡
Assume that 𝐹 is uniform,
𝑃 =
𝐹 ∙ 𝑑𝑟
𝑑𝑡
= 𝐹 ∙ 𝑣
Energy
;the capacity to do work
Mechanical Energy
2 types;
Potential Energy & Kinetic Energy
Potential Energy
(i) Elastic Potential Energy, 𝑈𝑠
For spring,
𝑊 1 →(2) =
1
2
𝑘𝑥2
2
−
1
2
𝑘𝑥1
2
(ii) Gravitational Potential Energy, 𝑈𝑔
An object initially at 𝑦𝑎 is lifted up by
𝐹 𝑎𝑝𝑝 to 𝑦 𝑏.
𝑊 = 𝐹 𝑎𝑝𝑝 ∙ ∆𝑟 = 𝑚𝑔𝑗 ∙ [(𝑦 𝑏 − 𝑦𝑎)𝑗]
𝑊 = 𝑚𝑔𝑦 𝑏 − 𝑚𝑔𝑦𝑎
𝑊 = ∆𝑈𝑔
𝑈𝑔 = 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦
***𝑈𝑔 depends on only vertical force(s).
Kinetic Energy
𝑤 = 𝐹 ∙ 𝑑𝑥
𝑥 𝑓
𝑥 𝑖
𝑃 = 𝐹 ∙ 𝑣
∆𝑟
𝑦𝑎
𝑦 𝑏
𝐹 𝑎𝑝𝑝
𝑚𝑔
0
𝑦
𝑥
𝑈𝑠 =
1
2
𝑘𝑥2
𝑈𝑔 = 𝑚𝑔𝑦
𝐹
∆𝑥
𝑣𝑖 𝑣 𝑓
8. Collisions
Law Of Conservation Of Momentum
𝑃𝑖 = 𝑃𝑓
***The law is applicable for all
collisions. Why?
; the Law is based on the 3rd law of
Newton, reaction and action forces.
According to the 3rd law of Newton,
𝐹12 = −𝐹21
𝐹12 + 𝐹21 = 0
According to the 2rd law of Newton,
𝑚1 𝑎1 + 𝑚2 𝑎2 = 0
𝑚1
𝑑𝑣1
𝑑𝑡
+ 𝑚2
𝑑𝑣2
𝑑𝑡
= 0
𝑑𝑚1 𝑣1
𝑑𝑡
+
𝑑𝑚2 𝑣2
𝑑𝑡
= 0
𝑑(𝑚1 𝑣1 + 𝑚2 𝑣2)
𝑑𝑡
= 0
𝐹12
𝑚1
𝑚2
𝐹21
𝑣1 𝑖
𝑣2 𝑖
𝑣1 𝑓
𝑣2 𝑓
𝑑(𝑃1 + 𝑃2)
𝑑𝑡
= 0
∴ 𝑃1 + 𝑃2 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃𝑖 = 𝑃𝑓
1 dimension elastic collision
From, 𝑃𝑖 = 𝑃𝑓 and 𝐸𝑖 = 𝐸𝑓, we can
derive another formula.
or
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ
=
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛
***The question will state whether the
collision is elastic or inelastic. However,
if the 2 masses stick together after the
collision, it is inelastic collision.
Elastic Inelastic
𝑃𝑖 = 𝑃𝑓
𝐸𝑖 = 𝐸𝑓
(Work due to non-
conservative force = 0)
𝐸𝑖 ≠ 𝐸𝑓
(Work due to non-
conservative force ≠ 0)
𝑣1𝑖
+ 𝑣1 𝑓
= 𝑣2 𝑖
+ 𝑣2 𝑓
𝑣1𝑖
− 𝑣2 𝑖
= 𝑣2 𝑓
− 𝑣1 𝑓