Power
Work In
Machines make work easier
Fulcrum
Work In = Work Out + Heat
Efficiency =
Work Out ! 100 Work In
Distance Out
Distance In
In a real machine the efficiency must always be less than 100%
Work In = Work Out Force x
Distance = Force x
Distance
Work Out
25 lbs 25 lbs
50 lbs
25 lbs 50 lbs
50 lbs
Give me a place to stand, and I’ll move the world! Archimedes
50 lbs
Machines make work easier.
He lifts the package up 4 ft…
Power 4 ft
4 ft
Power: the rate at which work is done
P= 8 ft
W F!d v = = F! v !t "t
4 ft
But the work stays the same. Remember…W = F x d
1
Units in the SI system….
A bulldozer and a small boy can do the same amount of work. It’s just that the bulldozer can do it faster.
Units in the English system
Power =
ft • lbs sec
550 ft ! lbs = 1 horsepower sec 1 HP = 746 Watts
Power =
Work time
Power =
Joules sec
= Watts
Example: A 2.00x103 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. Assume that the force of air resistance remains constant at a value of –500. N during this time
Find
the average power developed by the engine (express in watts and hp). Find the instantaneous power when the car reaches its final velocity (in watts and hp).
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. Assume that the force of air resistance remains constant at a value of –500. N during this time.
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time.
A) Find the average power developed by the engine (express in watts and hp). v v v v vf = vi + at vi = 0 m/s v 20.0 m/s = (a)(15.0 s) vf = 20.0 m/s
A) Find the average power developed by the engine (express in watts and hp). Feng = 3160 N v + vi v vavg = f 2 20.0 m/s + 0 m/s v vavg = 2 v vavg = 10.0 m/s
t = 15.0 s
a = 1.33m/s2
v
# F = Fengine " Fair resistance = m ! a
Feng " 500. N = (2000. kg)(1.33m/s 2 ) Feng = 3160 N
v 4 P = F ! vavg = (3160 N)(10.0 m/s) = 3.16 ! 10 W P = 3.16 ! 104 W !
1 hp = 42.4 hp 746 W
!
2
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time.
B) Find the instantaneous power when the car reaches its final velocity (in watts and hp).
P = F ! vf P = (3160 N)(20.0 m/s)
P = 6.32 ! 104 W P = 6.32 ! 104 W !
1 hp = 84.8 hp 746 W
Example: An elevator has a mass of 1000. kg and carries a maximum load of 800. kg. A constant frictional force of 4000. N retards its motion upward. What minimum power must the motor deliver to lift the fully-loaded elevator at a constant speed of 3.00 m/s? (In watts and horsepower)
v is constant, so a=0 FT – Ff – mg = 0 FT = Ff + mg = 4000 N + (1800 kg)(9.8m/s2 ) FT = 2.16 x 104 N P = FT v P = (2.16 x 104 N)(3.00 m/s) = 6.48 x 104 W P = (6.48 x 104 W)(1 hp / 746 W) = 86.9 hp
3
Machines make work easier
Fulcrum
Work In = Work Out + Heat
Efficiency =
Work Out ! 100 Work In
Distance Out
Distance In
In a real machine the efficiency must always be less than 100%
Work In = Work Out Force x
Distance = Force x
Distance
Work Out
25 lbs 25 lbs
50 lbs
25 lbs 50 lbs
50 lbs
Give me a place to stand, and I’ll move the world! Archimedes
50 lbs
Machines make work easier.
He lifts the package up 4 ft…
Power 4 ft
4 ft
Power: the rate at which work is done
P= 8 ft
W F!d v = = F! v !t "t
4 ft
But the work stays the same. Remember…W = F x d
1
Units in the SI system….
A bulldozer and a small boy can do the same amount of work. It’s just that the bulldozer can do it faster.
Units in the English system
Power =
ft • lbs sec
550 ft ! lbs = 1 horsepower sec 1 HP = 746 Watts
Power =
Work time
Power =
Joules sec
= Watts
Example: A 2.00x103 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. Assume that the force of air resistance remains constant at a value of –500. N during this time
Find
the average power developed by the engine (express in watts and hp). Find the instantaneous power when the car reaches its final velocity (in watts and hp).
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. Assume that the force of air resistance remains constant at a value of –500. N during this time.
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time.
A) Find the average power developed by the engine (express in watts and hp). v v v v vf = vi + at vi = 0 m/s v 20.0 m/s = (a)(15.0 s) vf = 20.0 m/s
A) Find the average power developed by the engine (express in watts and hp). Feng = 3160 N v + vi v vavg = f 2 20.0 m/s + 0 m/s v vavg = 2 v vavg = 10.0 m/s
t = 15.0 s
a = 1.33m/s2
v
# F = Fengine " Fair resistance = m ! a
Feng " 500. N = (2000. kg)(1.33m/s 2 ) Feng = 3160 N
v 4 P = F ! vavg = (3160 N)(10.0 m/s) = 3.16 ! 10 W P = 3.16 ! 104 W !
1 hp = 42.4 hp 746 W
!
2
Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time.
B) Find the instantaneous power when the car reaches its final velocity (in watts and hp).
P = F ! vf P = (3160 N)(20.0 m/s)
P = 6.32 ! 104 W P = 6.32 ! 104 W !
1 hp = 84.8 hp 746 W
Example: An elevator has a mass of 1000. kg and carries a maximum load of 800. kg. A constant frictional force of 4000. N retards its motion upward. What minimum power must the motor deliver to lift the fully-loaded elevator at a constant speed of 3.00 m/s? (In watts and horsepower)
v is constant, so a=0 FT – Ff – mg = 0 FT = Ff + mg = 4000 N + (1800 kg)(9.8m/s2 ) FT = 2.16 x 104 N P = FT v P = (2.16 x 104 N)(3.00 m/s) = 6.48 x 104 W P = (6.48 x 104 W)(1 hp / 746 W) = 86.9 hp
3
