Constants
𝑁𝑚2
Gravitational constant (g) = 6.67 x 10-11 𝑘𝑔2
𝑁𝑚2
Coulomb’s constant (k)= 9 x 109 𝐶2
Electron mass (me)= 9.11 x 10-31 kg
Proton mass (mp) = 1.67 x 10-27 kg
Elementary charge= ± 1.6 x 10-19 C
eV (electron volt)= 1.6 x 10-19 J
Planck’s constant (h)= 6.626 x 10-34 Js
𝑚
g(acceleration due to gravity)= 9.8
𝑠2
Units and Legends
Force= N
Energy= J
Momentume= kg m/s
Unit 1
Equations Of Motion
• Vf2= Vi2 + 2aΔd
• Vf= vi+at
• Δd= vit + ½ at2
• Δd= vft- ½ at2
𝑣1+𝑣2
• Δd= 𝑡
2
𝑣𝑓−𝑣𝑖
• A= 𝑡
𝑑𝑥 (𝑟𝑎𝑛𝑔𝑒)
• Vx=
𝑡
Newton law of motions
• Fg= mg
• FN= Fgy
Friction
• Ff= μFk/μFs (FN)
Ff
• =μ
FN
• μ (coefficient of friction) = angle
, Frequency and period
• (f) = number of cycles /second
• RPM = 60f
• f = RPM/60
Centripetal acceleration and centripetal force formula
• Ac= 4π2rf2
4π2 r
• Ac= 𝑡2
𝑚𝑣 2
• Ac= 𝑟
• Fc= Mac
2πr
• v= 𝑇
• v=2πrf
Finding acceleration in a pulley
• Fnet1= m1a
• Fnet2=m2a
• Fnet1= Fg1-Ft1
• Fnet 2=FT-Fg 2
• a=Fnet 1+Fnet 2
Finding acceleration of object with unspecified mass in a pulley
a=gsinθ1-gsinθ2
Finding FN of an elevator
Travelling upwards Travelling at constant speed
ma=FN-Fg FN=Fg
Travelling downwards
ma=Fg-FN
Finding applied force based on tension on an angle
• Fnet=Ftsinθ+Fg+Fk
• Max tension= ma+mg
• Min tension= ma-mg
Unit 2
Work done by friction
FF d (cos180)
Hooke’s law
• Fx= k|x|
• EE =½ kx2
𝑁𝑚2
Gravitational constant (g) = 6.67 x 10-11 𝑘𝑔2
𝑁𝑚2
Coulomb’s constant (k)= 9 x 109 𝐶2
Electron mass (me)= 9.11 x 10-31 kg
Proton mass (mp) = 1.67 x 10-27 kg
Elementary charge= ± 1.6 x 10-19 C
eV (electron volt)= 1.6 x 10-19 J
Planck’s constant (h)= 6.626 x 10-34 Js
𝑚
g(acceleration due to gravity)= 9.8
𝑠2
Units and Legends
Force= N
Energy= J
Momentume= kg m/s
Unit 1
Equations Of Motion
• Vf2= Vi2 + 2aΔd
• Vf= vi+at
• Δd= vit + ½ at2
• Δd= vft- ½ at2
𝑣1+𝑣2
• Δd= 𝑡
2
𝑣𝑓−𝑣𝑖
• A= 𝑡
𝑑𝑥 (𝑟𝑎𝑛𝑔𝑒)
• Vx=
𝑡
Newton law of motions
• Fg= mg
• FN= Fgy
Friction
• Ff= μFk/μFs (FN)
Ff
• =μ
FN
• μ (coefficient of friction) = angle
, Frequency and period
• (f) = number of cycles /second
• RPM = 60f
• f = RPM/60
Centripetal acceleration and centripetal force formula
• Ac= 4π2rf2
4π2 r
• Ac= 𝑡2
𝑚𝑣 2
• Ac= 𝑟
• Fc= Mac
2πr
• v= 𝑇
• v=2πrf
Finding acceleration in a pulley
• Fnet1= m1a
• Fnet2=m2a
• Fnet1= Fg1-Ft1
• Fnet 2=FT-Fg 2
• a=Fnet 1+Fnet 2
Finding acceleration of object with unspecified mass in a pulley
a=gsinθ1-gsinθ2
Finding FN of an elevator
Travelling upwards Travelling at constant speed
ma=FN-Fg FN=Fg
Travelling downwards
ma=Fg-FN
Finding applied force based on tension on an angle
• Fnet=Ftsinθ+Fg+Fk
• Max tension= ma+mg
• Min tension= ma-mg
Unit 2
Work done by friction
FF d (cos180)
Hooke’s law
• Fx= k|x|
• EE =½ kx2
