Data
F (your weight) =_________________________lb
Δy (height of staircase) =_________________________ft
time (each step) = _________________________s
time (skipping) = _________________________s
Calculations – for both runs, calculate
| 1) | Power in |
ft lb s |
by using P = | F Δy t |
2) Power in HP by using HP = P/550
3) Power in Watts by using W = HP × 746
Results – put all 6 answers into a results table
Question – Which run had the greatest Horsepower? Why?
Energy – the ability to do work. Units are the same as for work (J).
Kinetic Energy – the energy of motion
EK = ½ m v² (m = mass v = velocity)
Example – A 100 kg car moving at 20 m/s. EK =?
EK = ½ (100) (20)² = 2.0 × 104 J
Potential Energy – the energy of location
Gravitational – depends on height
EG = m g h (m = mass g = 9.8 m/s² h = height)
(Or EG = w h) (w = weight)
Example – A 20 kg object is 10 m above the ground. EG =?
EG = (20) (9.8) (10) = 1960 = 2.0 × 103 J (relative to the ground)
*When an object is lifted, the EG gained is equal to the work done in lifting the object.
*Falling Objects – As an object falls, the EG it loses is exactly equal to the EK it gains (conservation of energy – total energy is constant)
*** Roller Coaster Problem
What will be the velocity of the ball at the top of the second hill (assume no friction)?
EG (1) = m g h = (5) (9.8) (10) = 490 J
EG (2) = m g h = (5) (9.8) (3) = 147 J
EG (lost) = EG (1) – EG (2) = 490 – 147 = 343 J (becomes EK)
EK (1) = ½ m v² = ½ (5) (4)² = 40 J
EK (2) = EK (1) + EG (lost) = 40 + 343 = 383 J
| EK = ½ m v² | 383 = ½ (5) v² | |
| 383 = 2.5 v² | ||
| v² = 153.2 | ||
| v = 1.2 × 10¹ m/s east |
Friction – a force that always opposes sliding. It occurs whenever two surfaces rub against each other.
FF = µFN
FF = the force of friction
µ(mu) = |
the coefficient of friction – a measure of the roughness of the two surfaces (no units) |
FN = the normal force = the force that is pushing away from the surface perpendicular to the object.
The force of friction is independent of speed and surface area.
µstart is always > µslide (or µstatic > µkinetic) It is harder to get an object started than to keep it moving.
Flat Horizontal Surface
 |
|
| FN = weight of object = mg | |
| *** FF = µFN = µmg | |
2nd Law Problems
Work against gravity – lifting
F – mg = ma (mg = weight of object)
Work against friction – sliding
F – µmg = ma (µmg = force of friction)
If the object moves at a constant speed, then the opposing forces cancel each other and a = 0.
Examples
1) How much force is needed to lift a 20 kg object at 3.0 m/s²?
| F – mg = ma | F – (20) (9.8) = (20) (3.0) | |
| F – 196 = 60 | ||
| F = 256 = 2.6 × 10² N up |
2) How much force is needed to slide a 30.0 kg object across a table at a constant speed if µ = 0.2?
| F – µmg = ma | F – (.2) (30) (9.8) = (30) (0) | |
| F = 58.8 = 5.88 × 10¹ N forward |
***Friction on an inclined plane (hill)
| F// = mg sin θ | FF = µ FN = µmg cos θ |
FA = applied force – from problem (may not exist)
F// – always downhill
FF – opposite the motion of the object
If object moves uphill use:
Fup – Fdown = ma
If object moves downhill use:
Fdown – Fup = ma