B) Kinetic Energy - the energy an object possesses due to its' motion
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Equation…..
| KE = (½)mv2 |
Relationships
KE/m
direct
| KE = (½)mv2 |
KE/v
direct square
Ex) A 60.0-kg runner has 1920 joules of kinetic
energy. At what speed is she running?
Ex) A 60.-kg runner has 1920 joules of kinetic energy. At what speed is she running?
| KE = (½)mv2 |
m = 60. kg
KE = 1920 J
V = ?
64. = V2
V = 8.0 m/s
Summary W, K E, PE
P.E. - Equivalent to work needed to get an object to where it is or into its present condition
Then ... ΔPE = 40 J
K.E. - equivalent to work needed to get an object to a certain speed or to stop it.
C) Power - rate of doing work or expending energy
| P | = | W/t | = | Fd/t | = | Fv |
| (watt) | = (J/s) | (N)m/s |
Scalar
Ex 1) A force of 40N is exerted on a mass, causing it to move with a constant speed of 5m/s. What power is used?
Ex 1) A force of 40N is exerted on a mass, causing it to move with a constant speed of 5m/s. What power is used?
| P | = | W/t | = | Fd/t | = | Fv |
| (watt) | = (J/s) | (N)m/s |
= Fv = (40 N)(5 m/s) =
200 Nm/s or 200 watts
Ex 2) A girl weighing 500. newtons takes 50. seconds to climb a flight of stairs 18 meters high. Her power output vertically is
Ex 2) A girl weighing 500. newtons takes 50. seconds to climb a flight of stairs 18 meters high. Her power output vertically is
|
F = 500 N t = 50 seconds d = 18 m P = ? |
|
| P | = | W/t | = | Fd/t | = | Fv |
| (watt) | = (J/s) | (N)m/s |
P = Fd/t
P = 500 N(18 m)/50 sec
P = 180 Nm/sec = 180 watts
Ex 3) A 95-kg student climbs 4.0 m up a rope in 3.0 seconds. What is the power output of the student?
Ex 3) A 95-kg student climbs 4.0 m up a rope in 3.0 seconds. What is the power output of the student?
|
m = 95 kg d = 4.0 m t = 3 sec P = ? |
|
| P | = | W/t | = | Fd/t | = | Fv |
| (watt) | = (J/s) | (N)m/s |
F = ma = 95 kg(10 m/s2)
= 950 N