Newton’s Laws of Motion. three laws of motion: fundamental laws of mechanics describe the motion of all macroscopic objects (i.e., everyday size objects) moving at ordinary speeds (i.e., much less than the speed of light) .
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three laws of motion: fundamental laws of mechanics
describe the motion of all macroscopicobjects (i.e., everyday size objects) moving at ordinary speeds (i.e., much less than the speed of light)
Among other accomplishments, Sir Isaac Newton (1642-1727) invented calculus, developed the laws of motion, and developed the law of gravitational attraction.
Top view of a person standing in the aisle of a bus. (A) The bus is at rest, and then starts to move forward. Inertia causes the person to remain in the original position, appearing to fall backward.
(B) The bus turns to the right, but inertia causes the person to retain the original straight line motion until forced in a new direction by the side of the bus.
F = mam = mass of the object
a = acceleration
F = net force acting on the object
If the force of tire friction (F1) and the force of air resistance (F2) have a vector sum that equals the applied force (Fa), the net force is zero. Therefore, the acceleration is zero (i.e., velocity is constant)
More mass results in lessacceleration when the same force is applied. With the same force applied, the riders and the bike with twice as much mass will have half the acceleration (with all other factors constant). Note that the second rider is not pedaling.
1N = 1 kgm/s2
1 lb = 1 slug x 1 ft/s2 (slug is the unit of mass in the English system)
W = mg
The amount of substance (matter) contained in an object
a scalar quantity (no direction)
metric unit: kg
the same everywhere in the universe
ex.: mass of an object on the Moon is the same as on the Earth
The force of gravity on an object
a vector ( direction: vertically down)
metric unit: N (English unit: lb)
calculated as: W = mg
(g = gravitational acceleration)
changes with location (with change in g)
on the Moon: gMoon = 1.6 m/s2
weight of an object on the Moon is about six times less than on the EarthA parallel between the mass and the weight of an object
Whenever two objects interact, the force exerted by the first object on second is equal in size and opposite in direction to the force exerted by the second object on the first.
F1 = F2
ac = v2/r
m = mass
v = velocity
r = radius of the circle
Fc = mv2/r
A ball is swung on the end of a string in a horizontal circle. The pulling force of the string on the ball acts as centripetal force and causes the ball to change direction continuously, or accelerate into a circular path. Without the unbalanced force acting on it, the ball would continue in a straight line.
F = G(m1m2)/d2
G = 6.67 x 10-11 Nm2/kg2 (proportionality constant)
The force of attraction (F) is proportional to the product of the masses (m1, m2) and inversely proportional to the square of the distance (d) between the centers of the two masses.
F = G (m1m2) / d2
W = F = G (mEm) / d2 mE = mass of the earth m= mass of the object d = distance from the center of the earth to the object
mg = G (mEm) / d2
g = G mE / d2
The force of gravitational attraction decreases inversely with the square of the distance from the center of the earth. Note the weight of a 70.0 kg person at various distances above the surface of the earth.
Gravitational attraction acts as a centripetal force that keeps the Moon from following the straight-line path shown by the dashed line to position A. It was pulled to position B by gravity and thus "fell" toward Earth the distance from the dashed line to B, resulting in a somewhat circular path.
Work = force x distance
Work = Fd
Mechanical energy is of two types:
Potential Energy (PE)
kinetic energy = 1/2 (mass) (velocity)2
KE = (1/2) mv2
unit of KE = kg (m/s)2 = kg m2/s2 =
= (kg m/s2) m =Nm = J
unit of KE = J
unit of KE = ft-lb
(A) Work is done on the bowling ball as a force (FB) moves it through a distance. (B) This gives the ball a kinetic energy equal to the amount of work done on it. (C) The ball does work on the pins and has enough remaining energy to crash into the wall behind the pins.
During a free fall an object :
PE lost = KE gained
mg(Δh) = (1/2)mv2 ; (initial velocity of the object is zero)
Δh= distance through which the object falls
Solving for v : v= [2g(Δh)]1/2
This allows you to calculate the velocity of a falling object using energy conservation principles.
The ball (mass = 1 kg) trades potential energy for kinetic energy as it falls. Notice that the ball had 98 J of potential energy when dropped and has a kinetic energy of 98 J just as it hits the ground (in absence of air resistance)
v = 0
v=(2g Δh)1/2 = (2 x 9.8 m/s2 x 5m)1/2 =9.9m/s
v=(2g Δh)1/2 =(2 x 9.8 m/s2 x 10 m)1/2 = 14 m/s
This pendulum bob loses potential energy (PE) and gains an equal amount of kinetic energy (KE) as it falls through as distance h. The process reverses as the bob moves up the other side of its swing.