work done (J) Energy Power
• the product of a force and the • principle of energy -derivations - ↳ rate of energy transferr
displace ment moved in the direction ↳ energy cannot be created or GPE=mgΔh 9=9-81ms"
of the force. destroyed only transferred from DE = W
W = foccos @ • work done is one form to the next. h DE = FOCCOSO = 1 p= ¥ = FE E EV
-
maximum when GPE = mgh cos (O)
⑨ 0=0. DE = W GPE = Mgh efficiency = useful output e
• 0 = 90. = 0J in E fina_=
I L initial = Work done KE-IM v2 total input ener
x
horizontal direction GPE=KE 2
DE-W
① V=%PE
mgh=½mv² V DE = foccoso
W = foccoso W-Max cos@ (no drag) 29h-v2 x DE = foc cos (O) = 1
5m 29h = 2×9.81×5 ΔE = foc
J-blgmts-2nd ↳ no unit
YE? = 9.9ms' ✓ 2=42+2 ax DE-Max
J-kgm²m-2 ✓ 2=202 4=0 DE - MI
GPE- max at start ¥-ax
EXAM QUESTIONS KE = Max at bottom KE = ½ MV2
I roller coaster (2018) ✓=
B power in motor-65kW ② DE-W mass-2kg
A → B in 32s GPE- KE - W
42m mass = 2600kg 5m Mgh-½mv²= foc
C
7m V8
A 12m
not using foccoso (2×9.81×5)-14×2×82) - Fx 7
① WE F0C as resistive force
D=#= FI 65×103%552 is along the same 34.1=77
plane F- 34.1 =←
4.9N
32×65×103 = 20800005 7 ↳ resistive
= 2MJ ✓ force
ii efficiency = useful output energy ✗ 100
total input energy
input energy = 2×1065
DE - W useful - G PE
Ev-Ey = W
Ev- COPE = Mgh = 2600×9.81×42
= 1071252
1071252 ✗ 100 = 53.6%
2×106 = 54% ✓
b) x - 36m B → C
resistive force = 2-8×103 N
B- GPE - mgh W-DE
C = KE =½MV² Foc = Mgh-½MV²
• the product of a force and the • principle of energy -derivations - ↳ rate of energy transferr
displace ment moved in the direction ↳ energy cannot be created or GPE=mgΔh 9=9-81ms"
of the force. destroyed only transferred from DE = W
W = foccos @ • work done is one form to the next. h DE = FOCCOSO = 1 p= ¥ = FE E EV
-
maximum when GPE = mgh cos (O)
⑨ 0=0. DE = W GPE = Mgh efficiency = useful output e
• 0 = 90. = 0J in E fina_=
I L initial = Work done KE-IM v2 total input ener
x
horizontal direction GPE=KE 2
DE-W
① V=%PE
mgh=½mv² V DE = foccoso
W = foccoso W-Max cos@ (no drag) 29h-v2 x DE = foc cos (O) = 1
5m 29h = 2×9.81×5 ΔE = foc
J-blgmts-2nd ↳ no unit
YE? = 9.9ms' ✓ 2=42+2 ax DE-Max
J-kgm²m-2 ✓ 2=202 4=0 DE - MI
GPE- max at start ¥-ax
EXAM QUESTIONS KE = Max at bottom KE = ½ MV2
I roller coaster (2018) ✓=
B power in motor-65kW ② DE-W mass-2kg
A → B in 32s GPE- KE - W
42m mass = 2600kg 5m Mgh-½mv²= foc
C
7m V8
A 12m
not using foccoso (2×9.81×5)-14×2×82) - Fx 7
① WE F0C as resistive force
D=#= FI 65×103%552 is along the same 34.1=77
plane F- 34.1 =←
4.9N
32×65×103 = 20800005 7 ↳ resistive
= 2MJ ✓ force
ii efficiency = useful output energy ✗ 100
total input energy
input energy = 2×1065
DE - W useful - G PE
Ev-Ey = W
Ev- COPE = Mgh = 2600×9.81×42
= 1071252
1071252 ✗ 100 = 53.6%
2×106 = 54% ✓
b) x - 36m B → C
resistive force = 2-8×103 N
B- GPE - mgh W-DE
C = KE =½MV² Foc = Mgh-½MV²
