PHYSICS - INTRODUCTION

A scalar quantity is a quantity that is described by a magnitude that is a real number associated with a unit of measurement; examples of scalar quantities are: mass, temperature, volume, surface, density, pressure, electric charge, energy, wavelength

A vector quantity is a quantity that is described by a magnitude that is a real number associated with a unit of measurement, and a direction; examples of vector quantities are: force, acceleration, gravitational field, magnetic field, electric field, momentum, angular momentum

Distance is the difference between two points; distance is a scalar quantity, it has only magnitude; distance is only positive

Displacement requires the direction and it is the difference between the initial position and the final position; displacement is a vector quantity, it has magnitude and direction; displacement can be positive or negative; by convention, the displacement is positive when it is east or north, and it is negative when it is west or south

Speed describes how fast something is moving; speed is a scalar quantity so it has only magnitude; speed is always positive

Velocity is like speed but it requires the direction; velocity is a vector quantity so it has magnitude and direction; velocity can be positive or negative; velocity is speed with direction

speed = |velocity|

d = v⋅t; distance = speed⋅time or displacement = velocity⋅time

d = v⋅t; v = d/t; t = d/v

d = (1/2)(v_{0}+v_{f})t

An object is moving at the speed of 50m/s; calculate the time needed to travel a distance of 1000m; d = v⋅t, t = d/v = (1000m)/(50m/s) = 20 s

average speed = distance/time; v = d/t; speed is associated with distance

average velocity = displacement/time; v = d/t; velocity is associated with displacement

An object moves 12m east and 20 meters west in 4 seconds; calculate the average speed and the average velocity; average speed = distance/time = (12m+20m)/(4s) = (32m)/(4s) = 8m/s; average velocity = displacement/time = (12m-20m)/(4s) = (-8m)/(4s) = -2m/s

Acceleration is the rate of change of the velocity of an object with respect to time and it is a vector quantity, it has magnitude and direction; acceleration is measured in metre per second squared, m/s^{2}

a = Δv/Δt = (v_{f}-v_{0})/t

v_{f} = v_{0}+a⋅t

d = v_{0}⋅t+(1/2)⋅a⋅t^{2}

v_{f}^{2} = v_{0}^{2}+2⋅a⋅d

a = (v_{f}^{2}-v_{0}^{2})/2⋅d; v_{f}^{2} = v_{0}^{2}+2⋅a⋅d, v_{f}^{2}-v_{0}^{2} = 2⋅a⋅d, a = (v_{f}^{2}-v_{0}^{2})/2⋅d

d = (v_{f}^{2}-v_{0}^{2})/2⋅a; v_{f}^{2} = v_{0}^{2}+2⋅a⋅d, v_{f}^{2}-v_{0}^{2} = 2⋅a⋅d, d = (v_{f}^{2}-v_{0}^{2})/2⋅a

When the acceleration is positive, the velocity is increasing; when the acceleration is negative, the velocity is decreasing

An object is moving with v_{0} = 12m/s and a = 4m/s^{2}; t = 0s, v_{f} = 12m/s; t = 1s, v_{f} = 16m/s; t = 2s, v_{f} = 20m/s; t = 3s, v_{f} = 24m/s; t = 4s, v_{f} = 28m/s

The speed increases if v = + and a = + or v = - and a = -; the speed decreases if v = + and a = - or v = - and a = +

An object is moving east at a speed of 24m/s, with an acceleration of -6m/s^{2}; time = 0s, velocity = 24m/s, speed = 24m/s; time = 1s, velocity = 18m/s, speed = 18m/s; time = 2s, velocity = 12m/s, speed = 12m/s; time = 3s, velocity = 6m/s, speed = 6m/s; time = 4s, velocity = 0m/s, speed = 0m/s; time = 5s, velocity = -6m/s, speed = 6m/s; time = 6s, velocity = -12m/s, speed = 12m/s; time = 7s, velocity = -18m/s, speed = 18m/s; speed = |velocity|

Gravitational acceleration of the planet Earth: -9.8m/s^{2}

Gravitational acceleration of the Moon: -1.6m/s^{2}

Gravitational acceleration acts only vertically; if a body is in motion it acts only on the y component of the velocity; g = a_{y} = -9.8m/s^{2}

Calculate the velocity of a vertically falling object, considering g = -9.8m/s^{2}; time = 0, velocity = 0m/s, speed = 0m/s; time = 1, velocity = -9.8m/s, speed = 9.8m/s; time = 2, velocity = -19.6m/s, speed = 19.6m/s; time = 3, velocity = -29.4m/s, speed = 29.4m/s

An object is thrown up at the speed of 29.4m/s and after it reaches its maximum height it is in free fall, considering g = -9.8m/s^{2}; t = 0, v_{y} = 29.4m/s; t = 1, v_{y} = 19.6m/s; t = 2, v_{y} = 9.8m/s; t = 3, v_{y} = 0m/s, the object has reached its maximum height; t = 4, v_{y} = -9.8m/s; t = 5, v_{y} = -19.6m/s; t = 6, v_{y} = -29.4m/s

To study the projectile motion and its trajectory, it is essential to consider horizontal and vertical velocity

Calculate the horizontal and the vertical velocity of a ball that is kicked off a cliff with horizontal velocity of 5m/s, considering g = -9.8m/s^{2}; a_{x} = 0, a_{y} = g = -9.8m/s^{2}; t = 0, v_{x} = 5m/s, v_{y} = 0m/s; t = 1, v_{x} = 5m/s, v_{y} = -9.8m/s; t = 2, v_{x} = 5m/s, v_{y} = -19.6m/s; t = 3, v_{x} = 5m/s, v_{y} = -29.4m/s

A ball is kicked in the air at an angle θ, it has a starting velocity, and then falls to the ground; v_{x} = v⋅cos(θ), v_{y} = v⋅sin(θ); gravity acts on v_{y}, gravity does not act on v_{x}; in this example v_{x} = 8m/s and v_{y} = 29.4m/s; t = 0, v_{x} = 8m/s, v_{y} = 29.4m/s; t = 1, v_{x} = 8m/s, v_{y} = 19.6m/s; t = 2, v_{x} = 8m/s, v_{y} = 9.8m/s; t = 3, v_{x} = 8m/s, v_{y} = 0m/s, the ball has reached its maximum height and from now on it begins to descend; t = 4, v_{x} = 8m/s, v_{y} = -9.8m/s; t = 5, v_{x} = 8m/s, v_{y} = -19.6m/s; t = 6, v_{x} = 8m/s, v_{y} = -29.4m/s; when the trajectory is symmetrical the speed is the same; v_{x} does not change in a typical projectile motion problem, only v_{y} changes according to the gravitational acceleration

The Newton's first law, also called principle of inertia, states that an object at rest will stay at rest, and an object in motion will stay in motion unless acted on by a net external force, that is if the net force on an object is zero, then the velocity of the object is constant

The Newton's first law is described by the formula F = m⋅a, where F is the net force applied, m is the mass of the body, and a is the body's acceleration

The Newton's third law states that all forces between two objects exist in equal magnitude and opposite direction; if one object A exerts a force F_{A} on a second object B, then B simultaneously exerts a force F_{B} on A, and the two forces are equal in magnitude and opposite in direction: F_{A} = −F_{B}

Newton's first law: an object at rest will remain at rest unless acted on by a net force, and an object in motion will continue in motion unless acted on by a net force

A force is a push or pull action; when a force pulls an object through a rope, it is a tension force

The only way to stop a moving object is to apply a force in the opposite direction

An object moving over a surface slows down and stops due to the force of friction; friction always opposes movement

According to the Newton's first law, an object in motion will continue in motion unless acted on by a force; if there is no friction the object will continue to move forever; friction slows objects down until they stop

Newton's second law: F = m⋅a; ΣF_{x} = m⋅a_{x}, ΣF_{y} = m⋅a_{y}, F_{net} = m⋅a

1N = (1kg⋅1m)/(1s^{2}); a newton is defined as the force which gives a mass of 1 kilogram an acceleration of 1 metre per second, per second, 1 kg⋅m/s2

A force of 80N pushes an object of 10kg, calculate the acceleration of the object; F = m⋅a, a = F/m = 80N/10kg = 8N/kg = 8 m/s^{2}; as long as the force is applied, the object will increase its speed by 8m/s every second; t = 0s, v = 0m/s; t = 1s, v = 8m/s; t = 2s, v = 16m/s; t = 3, v = 24m/s

If a force is applied to an object, the object will accelerate and its speed will increase; if the force opposes the movement, the object will decelerate and its speed will decrease

Work is the energy transferred to or from an object via the application of force along a displacement; work is often represented as the product of force and displacement; a force is said to do positive work if it has a component in the direction of the displacement of the point of application; a force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force; when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball, that is a force, multiplied by the distance to the ground, that is a displacement; Work = Force⋅displacement; when the force F is constant and the angle between the force and the displacement d is θ, then the work done is W = F⋅d⋅cos(θ); work is a scalar quantity, so it has only magnitude and no direction; work transfers energy from one place to another, or one form to another; the unit of measurement of work in the International System of Units is the joule, the same unit as for energy

1J = 1N⋅1m = 1kg⋅(1m/s^{2})⋅1m = (1kg⋅1m^{2})(1s^{2}) = 1W⋅1s

When a force acts on an object and moves it by some displacement d, the force does a work on this object that is W = F⋅d; F and d are two vectors and W = F⋅d is valid when the vector F is parallel to the vector d, so in general W = F⋅d⋅cos(θ)

The work accomplished by the force is the product of the magnitude of the force times its displacement

Energy is the quantitative property that must be transferred to a body or physical system to perform work on the body, or to heat it; energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed; the unit of measurement of energy in the International System of Units is the joule, which is the energy transferred to an object by the work of moving it a distance of one metre against a force of one newton

Energy is the ability to do work; an object with energy has the ability to do work; when a force acts on an object, the force transfers energy to the object

Kinetic Energy = (1/2)⋅mass⋅velocity^{2}; KE = (1/2)⋅m⋅v^{2}; kinetic energy in J, mass in kg, velocity in m/s

Potential Energy = mass⋅gravitational acceleration⋅height; PE = m⋅g⋅h; potential energy in J, mass in kg, gravitational acceleration im m/s^{2}, height in m

A moving object has kinetic energy; a stationary object has no kinetic energy

W_{net} = ∆KE, it is the work energy theorem; when a force increases the kinetic energy of an object, the force does a positive work on the object and the object speeds up; when a force decreases the kinetic energy of an object, the force does a negative work on the object and the object slows down

W_{net} = ∆KE = KE_{f}-KE_{0} = (1/2)⋅m⋅v_{f}^{2}-(1/2)⋅m⋅v_{0}^{2} = m((1/2)⋅v_{f}^{2}-(1/2)⋅v_{0}^{2}), W = F⋅d, F = m⋅a, W = m⋅a⋅d, m((1/2)⋅v_{f}^{2}-(1/2)⋅v_{0}^{2}) = m⋅a⋅d, (1/2)⋅v_{f}^{2}-(1/2)⋅v_{0}^{2} = a⋅d, (1/2)(v_{f}^{2}-v_{0}^{2}) = a⋅d, v_{f}^{2}-v_{0}^{2} = 2⋅a⋅d, v_{f}^{2} = v_{0}^{2}+2⋅a⋅d

A moving object hits a stationary object, therefore the moving object exerts a force, called action force, on the stationary object, so the moving object performs work on the stationary object, but at the same time the stationary object performs work on the moving object, exerting an equal and opposite force, called reaction force; when there is a force acting on an object, there is a transfer of energy; the moving object hits the stationary object by exerting a force, therefore the moving object does a positive job on the stationary object by transferring a part of its kinetic energy; the stationary object exerts a reaction force on the moving object, decreasing its kinetic energy and slowing it down; when two objects collide, there is a transfer of energy; when the force and the displacement are two parallel vectors, the work done by the force on the object is positive because W = F⋅d⋅cos(0°) = 1; when the force and the displacement are opposite vectors, the work done by the force on the object is negative because W = F⋅d⋅cos(180°) = F⋅d⋅-1; if the force and the displacement vectors are perpendicular, the force does no work on the object because W = F⋅d⋅cos(90°) = F⋅d⋅0 = 0, but this force acts as a centripetal force; the action force vector and the displacement vector have the same direction, so the action force does a positive work, therefore the moving object hits the stationary object and makes it move; the reaction force vector and the displacement vector have opposite direction, so the reaction force does a negative work, therefore the object that is hit slows the moving object

Gravity is a downward force, it causes objects to move downward; gravity force acts in the negative y direction; if an object moves upward, the gravity force does a negative work, slowing the object, because the displacement vector of the object has opposite direction of the gravity force vector; when an object is falling, the work done by the gravity force is positive because the displacement vector of the falling object has the same direction of the gravity force vector; when an object is thrown into the air, its velocity decreases and its kinetic energy also decreases, ∆KE is negative, W = ∆KE, so the work is negative, the gravity force does a negative work on an object moving upward because the displacement vector of the object and the gravity force vector have opposite direction; the object reaches its maximum height and then begins to fall towards the ground, increasing its velocity, and during the descent the gravity force does a positive work on the falling object because the displacement vector of the object and the gravity force vector have the same direction; the object has a negative acceleration when it goes up, and it has a positive acceleration when it goes down; the acceleration is always in the direction of the net force; when the velocity vector and the acceleration vector have opposite direction, the object slows; the falling object has a force vector and therefore an acceleration vector directed downwards and the velocity vector also has the same downward direction, so the object increases the speed heading towards the ground; the kinetic energy of the falling object is positive, ∆KE is positive, so the work is positive, the gravity force does a positive work on the falling object; Potential Energy = m⋅g⋅h; when the object is thrown upwards it increases its potential energy, reaches the maximum height and therefore the maximum of its potential energy, then it begins to descend and its potential energy decreases until it reaches zero when the object reaches the ground; when the object goes up its kinetic energy is converted into potential energy, and when the object goes down its potential energy is converted into kinetic energy

Mechanical Energy = Potential Energy + Kinetic Energy, ME = PE+KE; mechanical energy is the sum of potential energy and kinetic energy; mechanical energy is conserved when the forces are conservative forces

Conservative forces: gravity, elastic force associated with springs, eletric force

Non-conservative forces: friction, air resistance, push and pull forces, tension force

An object is thrown into the air with a force F_{A}; the velocity vector and the displacement vector are directed upward but the gravitational force vector is directed downward; the resulting force vector is directed upward; the kinetic energy increases, the potential energy increases, the mechanical energy increases considering that ME = PE+KE; the work done by F_{A} is positive, the work done by the gravitational force is negative, the work done by the resulting force is positive

Power is the amount of energy transferred or converted per unit time; in the International System of Units, the unit of power is the watt, equal to one joule per second; power is a scalar quantity

Power is the rate at which energy is transferred

Power = Work/Time; P = W/t; 1Watt = 1Joule/1second; 1W = 1J/1s; 1kW = 1000W; 1MW = 1⋅10^{6}W; 1hp = 746W

Power = Force⋅Velocity; P = F⋅V; Power = Work/Time, Work = Force⋅Displacement, Power = Force⋅Displacement/Time, Displacement/Time = Velocity, Power = Force⋅Velocity

A machine exerts a force of 100 Newton for a displacement of 1 meter in 1 second, calculate the power; W = F⋅d = 100N⋅1m = 100J, P = W/t = 100J/1s = 100W

A machine exerts a force of 100 Newton for a displacement of 1 meter in 10 second, calculate the power; W = F⋅d = 100N⋅1m = 100J, P = W/t = 100J/10s = 10J/1s = 10W

A machine is more powerful than another if it can do the same work in less time

Calculate the kinetic energy of a 5kg block sliding across a frictionless horizontal surface at 12m/s; KE = (1/2)⋅m⋅v^{2} = (1/2)⋅5kg⋅(12m/s)^{2} = (1/2)⋅5kg⋅144m^{2}/s^{2} = 360kg⋅m^{2}/s^{2} = 360N⋅m = 360J

Calculate what happens to an object's kinetic energy if the mass is doubled; KE = (1/2)⋅m⋅v^{2}; KE_{1} = (1/2)⋅1⋅1^{2} = 1/2; KE_{2} = (1/2)⋅2⋅1^{2} = 1; if the mass is doubled the kinetic energy is doubled

Calculate what happens to an object's kinetic energy if the speed is doubled; KE = (1/2)⋅m⋅v^{2}; KE_{1} = (1/2)⋅1⋅1^{2} = 1/2; KE_{2} = (1/2)⋅1⋅2^{2} = 2; if the speed is doubled the kinetic energy is quadrupled

Calculate what happens to an object's kinetic energy if the speed is tripled; KE = (1/2)⋅m⋅v^{2}; KE_{1} = (1/2)⋅1⋅1^{2} = 1/2; KE_{2} = (1/2)⋅1⋅3^{2} = 9/2; if the speed is tripled the kinetic energy is multiplied by 9

Calculate what happens to an object's kinetic energy if the mass is tripled and the speed is quadrupled; KE = (1/2)⋅m⋅v^{2}; KE_{1} = (1/2)⋅1⋅1^{2} = 1/2; KE_{2} = (1/2)⋅3⋅4^{2} = 48/2; if the mass is tripled and the speed is quadrupled the kinetic energy is multiplied by 48

Calculate the gravitational potential energy of a 2.5kg book that is 10m above the ground; PE = m⋅g⋅h = 2.5kg⋅9.8m/s^{2}⋅10m = 245kg⋅m^{2}/s^{2} = 245N⋅m = 245J

A 10kg ball falls from a height of 100m, calculate the vertical speed of the ball during the first 4 seconds, calculte the height of the ball above the ground during the first 4 seconds, calculate the kinetic and potential energies of the ball, determine the total mechanical energy of the ball, explain if gravity is a conservative force; v_{f} = v_{0}+a⋅t, d = v_{0}⋅t+(1/2)⋅a⋅t^{2}, PE = m⋅g⋅h, KE = (1/2)⋅m⋅v^{2}, ME = PE+KE; t = 0s, v = (9.8m/s^{2})⋅0s = 0m/s, h = 100m, PE = 10kg⋅9.8m/s^{2}⋅100m = 9800J, KE = 0J, ME = 9800J+0 = 9800J; t = 1s, v = (9.8m/s^{2})⋅1s = 9.8m/s, h = 100m-(1/2)⋅(9.8m/s^{2})⋅(1s)^{2} = 95.1m, PE = 10kg⋅9.8m/s^{2}⋅95.1m = 9319.8J, KE = (1/2)⋅10kg⋅(9.8m/s)^{2} = 480.2J, ME = PE+KE = 9319.8J+480.2J = 9800J; t = 2s, v = (9.8m/s^{2})⋅2s = 19.6m/s, h = 100m-(1/2)⋅(9.8m/s^{2})⋅(2s)^{2} = 80.4m, PE = 10kg⋅(9.8m/s^{2})⋅80.4m = 7879.2J, KE = (1/2)⋅10⋅(19.6)^{2} = 1920.8J = ME = 7879.2J+1920.8J = 9800J; t = 3s, v = (9.8m/s^{2})⋅3s = 29.4m/s, h = 100-(1/2)⋅(9.8m/s^{2})⋅(3s)^{2} = 55.9m, PE = 10kg⋅(9.8m/s^{2})⋅55.9m = 5478.2J, KE = (1/2)⋅10kg⋅(29.4m/s)^{2} = 4321.8J, ME = 5478.2J+4321.8J = 9800J; t = 4s, v = (9.8m/s)⋅4s = 39.2m/s , h = 100-(1/2)⋅(9.8m/s^{2})⋅(4s)^{2} = 21.6m, PE = 10kg⋅(9.8m/s^{2})⋅21.6m = 2116.8J; KE = (1/2)⋅10kg⋅(39.2m/s)^{2} = 7683.2J, ME = 2116.8J+7683.2J = 9800J; mechanical energy is conserved, so gravity is a conservative force

A 70N force is applied horizontally to a 10kg block at rest for a displacement of 200m across a frictionless surface, calculate the work done by the force, the final kinetic energy, the speed, the horizontal acceleration, and use kinematics to calculate the final speed; W = F⋅d = 70N⋅200m = 14000J; W = ΔE = KE_{f}-KE_{0} = KE_{f} = 14000J; W = (1/2)⋅m⋅v^{2}, 14000J = (1/2)⋅10⋅v^{2}, v = √14000/5J = 52.9J; F = m⋅a, a = F/m = 70N/10kg = 7m/s^{2}; v_{f}^{2} = v_{0}^{2}+2⋅a⋅d, v_{f} = √2⋅(7m/s^{2})⋅200m = √2800m^{2}/s^{2} = 52.9m/s

Calculate how much work is required to accelerate a 1500kg car from 15m/s to 40m/s, and calculate the average net force acting on the car if it reaches a final speed of 40m/s while traveling a distance of 275m; W = ∆E = KE_{f}-KE_{0} = (1/2)⋅m⋅v_{f}^{2}-(1/2)⋅m⋅v_{0}^{2} = (1/2)⋅m(v_{f}^{2}-v_{0}^{2}) = (1/2)⋅1500kg((40m/s)^{2}-(15m/s)^{2}) = 1031250J; W = F⋅d, F = W/d = 1031250J/275m = 3750N; a = (v_{f}^{2}-v_{0}^{2})/2d = ((40m/s)^{2}-(15m/s)^{2})/2⋅275m = 2.5m/s^{2}; F = m⋅a = 1500kg⋅2.5m/s^{2} = 3750N

Calculate how much work is done by a constant 50N force that acts over a displacement of 10m, and how much work is done by a varying force that increases at a constant rate from 40N to 80N over a displacement of 10m; W = F⋅d = 50N⋅10m = 500J; W = F_{average}⋅d = (1/2)(F_{0}+F_{f})⋅d = (1/2)(40+80)10 = 600J

A car travels 200 miles east in 4 hours and then 300 miles west in 5 hours, calculate the average speed of the car for the entire trip; speed = distance/time = (200mi+300mi)/(4hrs+5hrs) = 500mi/9hrs = 55.6mph

A car travels 200 miles east in 4 hours and then 300 miles west in 5 hours, calculate the average velocity of the car for the entire trip; distance and speed are scalar quantities, they have only magnitude and they can be only positive; displacement and velocity are vectors, they have magnitude and direction and they can be positive or negative; velocity = displacement/time = (200mi-300mi)/(4hrs+5hrs) = -100mi/9hrs = -11.1mph

A car accelerates from 15m/s to 45m/s in 9s, calculate how far the car travel; d = (1/2)(v_{0}+v_{f})t = (1/2)(15m/s+45m/s)9s = (1/2)(60m/s)9s = 270m

A ball in projectile motion is kicked horizontally off a 200m cliff at an initial speed of 15m/s; the initial vertical velocity is zero, true; the horizontal acceleration is -9.8m/s^{2}, false; the horizontal speed is 15m/s when it hits the ground, true; the vertical velocity decreases by 9.8m/s every seconds, true; the horizontal speed is constant, true; usually in a typical projectile motion problem v_{x} = constant, a_{x} = 0, a_{y} = -9.8m/s^{2}

A ball is released from rest at the top of a 450m cliff, calculate the time it takes to reach the ground; d = v_{0}⋅t+(1/2)⋅a⋅t^{2} = (1/2)⋅a⋅t^{2}, t^{2} = 2d/a, t = √2d/a = √(2⋅450m)/(9.8m/s^{2}) = √(900m)/(9.8m/s^{2}) = √91.84s^{2} = 9.58s

A ball is thrown downward with an initial speed of 25m/s from the top of a 500m building, calculate the time it takes to reach the ground; d = v_{0}⋅t+(1/2)⋅a⋅t^{2}, (1/2)⋅a⋅t^{2}+v_{0}⋅t-d = 0, (1/2)(9.8m/s^{2})t^{2}+(25m/s)t-500m = 0, a = 4.9, b = 25, c = -500, t = (-b±√b^{2}-4ac)/2a, t = (-25±√25^{2}-4⋅4.9(-500))/(2⋅4.9) = (-25±√625+9800)/9.8 = (-25±√10425)/9.8 = (-25±102.1)/9.8 = 77.1/9.8 = 7.87s, the negative solution cannot be considered

The velocity of a rock in component form is v = -12i+5j, calculate the speed of the rock and its direction with respect to the +x axis; v_{x} = -12m/s, v_{y} = 5m/s; the speed is the magnitude of the velocity that is the hypotenuse of a right-angled triangle calculable with the Pythagorean theorem v = √v_{x}^{2}+v_{y}^{2} = √(-12m/s)^{2}+(5m/s)^{2} = √(144m^{2}/s^{2})+(25m^{2}/s^{2}) = √169m^{2}/s^{2} = 13m/s; α = arctan(5/12) = arcsin(5/13) = arccos(12/13) = 22.6°, but the velocity vector is in the second quadrant so the direction with respect to the +x axis is 180°-22.6° = 157.4°

A ball rolls horizontally off a cliff at 23m/s, and the range of the ball is 184m, calculate the height of the cliff; d = vt where d is the range, t = d/v = (184m)/(23m/s) = 8s; d = v_{0}t+(1/2)a⋅t^{2} where d is the height, v_{0} = 0, h = (1/2)(9.8m/s^{2})(8s)^{2} = 313.6m

A truck slowly speeds up from 15km/h to 95km/h in 2min, calculate the average acceleration of the truck in m/s^{2}; v_{0} = 15km/h = 15(1000/3600)m/s = 4.167m/s, v_{f} = 95km/h = 95(1000/3600)m/s = 26.389m/s, t = 2min = 120s, a = Δv/Δt = v_{f}-v_{0}/t = (26.389m/s-4.167m/s)/120s = (22.222m/s)/120s = 0.185m/s^{2}

An inclined plane, also known as a ramp, is a flat supporting surface tilted at an angle, with one end higher than the other, used as an aid for raising or lowering a load; moving an object up an inclined plane requires less force than lifting it straight up, at a cost of an increase in the distance moved; the mechanical advantage of an inclined plane depends on its slope, meaning its gradient or steepness; the smaller the slope, the larger the mechanical advantage, and the smaller the force needed to raise a given weight; a plane's slope s is equal to the difference in height between its two ends, or rise, divided by its horizontal length, or run, and it can also be expressed by the angle θ that the plane makes with the horizontal; θ = arctan(rise/run) = arcsin(rise/slope) = arccos(run/slope); a = g⋅sin(θ)

Calculate the acceleration of a block on an inclined plane with angle θ; F_{n} = m⋅g⋅cos(θ), F_{g} = m⋅g⋅sin(θ), ΣF_{x} = F_{g}, m⋅a = m⋅g⋅sin(θ), a = g⋅sin(θ)

An 8kg box lies on a 30 degree frictionless inclined plane, so calculate the final speed of the block as it slides down the inclined plane starting from the rest for a distance of 200m; a = g⋅sin(θ) = 9.8m/s^{2}⋅sin(30°) = 9.8m/s^{2}⋅(1/2) = 4.9m/s^{2}; v_{f}^{2} = V_{0}^{2}+2⋅a⋅d = 0+2⋅4.9m/s^{2}⋅200m = 9.8m/s^{2}⋅200m = 1960m^{2}/s^{2}, V_{0} = √1960m^{2}/s^{2} = 44.27m/s

A tension force of 500N acts on a 20kg block pulling it to the right, and a constant kinetic frictional force of 140N acts on the block, so calculate the acceleration of the block; friction always opposes motion; ΣF_{x} = (Tension Force)-(Frictional Force) = T_{f}-F_{f} = 500N-140N = 360N; F = m⋅a, a = F/m = 360N/20kg = 18m/s^{2}

An object is moving in a straight line in the +x direction with constant speed; the net force acting on the object is zero, true because only when the net force is zero the velocity is constant; the kinetic frictional force is zero, false because there may be a force opposed to friction such that the resulting acceleration is zero; the horizontal acceleration is zero, true because the velocity is constant; the normal force is zero, false because the normal force F_{N} = m⋅g

A 25kg block rests on a horizontal surface and a tension force of 135N is applied on the block in the +y direction to pull it up, so calculate the normal force acting on the block; weight force F_{W} = m⋅g = 25kg⋅9.8m/s^{2} = 245N; normal force = (weight force)-(tension force), F_{N} = F_{W}-F_{T} = 245N-135N = 110N

A car is located 350m west of town XYZ and is traveling east at 25m/s with a constant acceleration of 3m/s^{2} in the +x direction, so calculate the position of the car relative to town XYZ after 12 seconds; d = v_{0}⋅t+(1/2)⋅a⋅t^{2} = 25m/s⋅12s+(1/2)⋅(3m/s^{2})⋅(12s)^{2} = 300m+216m = 516m east from the starting point, so the car is 516m-350m = 166m east of the town XYZ

A ball is kicked at 25 m/s from the ground at a launch angle of 20 degrees, so calculate the other launch angle for the same range; 90°-20° = 70°; the maximum range of any projectile is obtained with a launch angle of 45°

Calculate how much work is required to speed up a 10kg block from rest to 16m/s; W = ΔKE = KE_{f}-KE_{0} = (1/2)⋅m⋅v^{2} = (1/2)(10kg)(16m/s)^{2} = 1280J

Calculate the work required to lift a 15kg rock 12 meters above the ground; W = ΔPE = mgΔh = (15kg)(9.8m/s^{2})(12m) = 1764J

One of these statements is associated with Newton's third law of translational motion; an oject at rest will remain at rest unless acted on by a net force, Newton's first law; the acceleration of an object is directly proportional to the net force acting on it, Newton's second law, F = m⋅a; the net force acting on an object is equal to the rate of change of the momentum of the object, Newton's second law, ΣF = m⋅a = (m⋅Δv)/Δt = centripetal accelerationP/Δt, P is the momentum, P = m⋅v; an object in motion will continue in motion unless acted on by a net force, Newton's first law; for every action force there is an equal and opposite reaction force which acts on different objects, Newton's third law

Ignoring friction and the inertia of the pulley, there are two blocks linked by a rope, the first block is on a surface and the second block is falling; the acceleration of the two blocks is (m_{2}⋅g)/(m_{1}+m_{2}), because F = m⋅a, a = F/m; ΣF_{y} = T-m_{2}⋅g, -m_{2}⋅a = T-m_{2}⋅g, m_{2}⋅g-m_{2}⋅a = T, ΣF_{x} = T, m_{1}⋅a = T, m_{2}⋅g-m_{2}⋅a = m_{1}⋅a, m_{2}⋅g = m_{1}⋅a+m_{2}⋅a, m_{2}⋅g = a(m_{1}+m_{2}), a = (m_{2}⋅g)/(m_{1}+m_{2})

A centripetal force is a force that makes a body follow a curved path; its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path; in Newtonian mechanics, gravity provides the centripetal force causing astronomical orbits; centripetal acceleration = Δv/Δt = v^{2}/r; centripetal force = F_{c} = m⋅a_{c} (m⋅v^{2})/r

A car travels around a curve of radius 500m at a speed of 15m/s, so calculate the centripetal acceleration of the car; A_{c} = v^{2}/r = (15m/s)^{2}/500m = 0.45m/s^{2}

A 5kg box slides on a horizontal surface with a coefficient of kinetic friction of 0.15 at an initial speed of 12 m/s, so calculate how far the box travel before coming to rest; ΣF_{x} = -F_{kf}, where F_{kf} is the force of kinetic friction, F_{kf} = μ_{k}F_{n}, m⋅a = -μ_{k}F_{n}, the normal force F_{n} = m⋅g, m⋅a = -μ_{k}⋅m⋅g, a = -μ_{k}⋅g = -0.15⋅9.8m/s^{2} = -1.47m/s^{2}; v_{f}^{2} = v_{0}^{2}+2⋅a⋅d, v_{f}^{2}-v_{0}^{2} = 2⋅a⋅d, d = (v_{f}^{2}-v_{0}^{2})/(2⋅a) = (0-(12m/s)^{2})/(2⋅(-1.47m/s^{2})) = (-144m^{2}/s^{2})/(-2.94m/s^{2}) = 48.98m

A 2kg ball attached to a 1.5m long string moves in a horizontal circle at 15m/s, so calculate the tension of the string; T_{x} = F_{c} = m⋅a_{c} = (m⋅v^{2})/r = 2kg⋅(15m/s)^{2}/1.5m = 300N; T_{y} = m⋅g = 2kg⋅9.8m/s^{2} = 19.6N; T = √T_{x}^{2}+T_{y}^{2} = √300N^{2}+19.6N^{2} = 300.6N

Calculate the maximum speed at which a car can safely round a curve of radius 400m, considering that the coefficient of static friction between the road and the tires is 0.2; F_{sf} is the force of static friction and F_{c} is the centripetal force, F_{sf} = F_{c}, μ_{s}F_{n} = m⋅v^{2}/r, μ_{s}⋅m⋅g = m⋅v^{2}/r, v^{2} = μ_{s}⋅r⋅g, v = √μ_{s}⋅r⋅g, √0.2⋅400m⋅9.8m/s^{2} = 28m/s

One of the following statements is not true concerning a ball that is thrown straight upward into the air: the speed of the ball is decreasing as the ball travels upward, true; the speed of the ball is increasing as the ball travels downward, true; the velocity of the ball is zero at the maximum height, true; the velocity of the ball is decreasing as the ball travels upward, true; the velocity of the ball is increasing as the ball travels downward, false

When the acceleration is positive the velocity increases; when the acceleration is negative the velocity decreases

When the acceleration and the velocity have the same sign, the speed increases; when the acceleration and the velocity have opposite sign, the speed decreases

The centripetal force always points towards the center of the circle

Calculate the effect on the kinetic energy when the speed doubles; KE = (1/2)⋅m⋅v^{2}; the kinetic energy increases by a factor of 2, false; the kinetic energy increases by a factor of 4, true; the kinetic energy will be 1/2 of its original value, false; the kinetic energy will be 1/4 of its original value, false; the kinetic energy will not change, false

Gravity is a natural phenomenon by which all things with mass or energy are attracted to one another; gravity is well approximated by Newton's law of universal gravitation, which describes gravity as a force causing any two bodies to be attracted toward each other, with magnitude proportional to the product of their masses and inversely proportional to the square of the distance between them; gravitational force F = G⋅m_{1}⋅m_{2}/r^{2}; gravitational constant G = 6.67⋅10^{-11}m^{3}/(kg⋅s^{2}) = 6.67⋅10^{-11}(N⋅m^{2})/kg^{2}

Two objects with masses m are separated by a distance r and the gravitational force between them is F, so find the expression that represents the gravitational force acting on each object if the masses are now 3m and 4m separated by a distance of (1/2)r; F = G⋅m_{1}⋅m_{2}/r^{2} = 1⋅3⋅4/(1/2)^{2} = 12/(1/4) = 12⋅4 = 48F

An object with mass m and speed v travels around a circle of radius r, and the centripetal force acting on this object is 500N, so calculate the new centripetal force acting on the object if the speed is doubled, the mass is tripled, and the radius of curvature is reduced to 1/2 of its original value; F_{c} = m⋅a_{c} = m⋅v^{2}/r; F_{c} = (3⋅(2)^{2}/(1/2))500N = (3⋅4⋅2)500N = 24⋅500N = 12000N

A satellite orbits the earth at 4500km above the surface of the planet, the mass of the earth is 5.98⋅10^{24}kg and the radius of the earth is 6.38⋅10^{6}m, so calculate the speed of the satellite; F_{g} = G⋅m_{e}⋅m_{s}/r^{2}; F_{c} = m_{s}⋅v_{s}^{2}/r; F_{g} = F_{c}, G⋅m_{e}⋅m_{s}/r^{2} = m_{s}⋅v_{s}^{2}/r, v_{s}^{2} = G⋅m_{e}/r, v_{s} = √G⋅m_{e}/r = √(6.67⋅10^{-11}(N⋅m^{2})/kg^{2})⋅(5.98⋅10^{24}kg)/(6.38⋅10^{6}m+4.5⋅10^{6}m) = 6054.8m/s

Elastic energy is the mechanical potential energy stored in the configuration of a material or physical system as it is subjected to elastic deformation by work performed upon it; elastic energy occurs when objects are impermanently compressed, stretched or generally deformed in any manner; the energy is potential as it will be converted into other forms of energy, such as kinetic energy and sound energy, when the object is allowed to return to its original shape by its elasticity

U = (1/2)k⋅∆x^{2}, U = elastic potential energy, k = spring constant, ∆x = length of spring compression

A horizontal spring with spring constant k = 500N/m is compressed 45cm by a 12kg block, so calculate the speed of the block as soon as it is released by the spring; the elastic potential energy is converted into kinetic energy, U = KE, (1/2)k⋅∆x^{2} = (1/2)⋅m⋅v^{2}, v^{2} = (k⋅∆x^{2})/m, v = √(k⋅∆x^{2})/m = √(500N/m⋅(0.45m)^{2})/12kg = 2.9m/s

A ball is kicked with a speed of 40m/s at an angle of 30° from a 500m cliff, so calculate the speed of the ball before it hits the ground; ME_{0} = ME_{f}, PE_{0}+KE_{0} = KE_{f}, m⋅g⋅h+(1/2)⋅m⋅v_{0}^{2} = (1/2)⋅m⋅v_{f}^{2}, g⋅h+(1/2)⋅v_{0}^{2} = (1/2)⋅v_{f}^{2}, v_{f}^{2} = 2(g⋅h+(1/2)⋅v_{0}^{2}), v_{f} = √2(g⋅h+(1/2)⋅v_{0}^{2}) = √2(9.8m/s^{2}⋅500m+(1/2)⋅(40m/s)^{2}) = 106.8m/s

Calculate the power exerted by the engine of a 1500kg car moving at a constant speed of 25m/s with a constant retarding force of 1200N working against it; the car is moving at a constant speed, so the acceleration is 0, and therefore the force exerted by the engine is equal to the retarding force; power = work/time, work = force⋅displacement, velocity = displacement/time, P = W/t = F⋅d/t = F⋅v = 1200N⋅25m/s = 30000W

A laptop uses 150W of power, and the cost of electricity is $0.06 per kWh in town XYZ, so calculate the cost of running the laptop 4 hours per day for an entire month; power = work/time, work = power⋅time, energy = power⋅time; kWh is a unit of energy; 150Wh:$x = 1000Wh:$0.06, $x = 150Wh⋅$0.06/1000Wh = $0.009 is the cost of electricity consumed by the laptop in one hour, $0.009⋅4 = $0.036 is the cost of electricity consumed by the laptop in four hours, $0.036⋅30 = $1.08 is the cost of electricity consumed by the laptop in 30 days with a use of 4 hours a day

In a graph where y = velocity and x = time, the slope of the graph at one point is the instantaneous acceleration; in a straight line the slope is constant, so the instantaneous acceleration is equal to the average acceleration; slope = rise/run, m = ∆y/∆x = (y_{2}-y_{1})/(x_{2}-x_{1}); a = ∆v/∆t = (v_{2}-v_{1})/(t_{2}-t_{1})

Impulse is the integral of a force F over the time interval t for which it acts; since force is a vector quantity, impulse is also a vector quantity; impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction; the SI unit of impulse is the newton second, N⋅s, and the dimensionally equivalent unit of momentum is the kilogram meter per second, kg⋅m/s

Impulse = (force)⋅(change in time), J = F⋅∆t

Linear momentum, translational momentum, or simply momentum, is the product of the mass and velocity of an object; it is a vector quantity, possessing a magnitude and a direction; if m is an object's mass and v is its velocity, then the object's momentum is p = m⋅v; in SI units, momentum is measured in kilogram meters per second, kg⋅m/s

Momentum = mass⋅velocity, p = m⋅v

Impulse-Momentum Formula: Impulse = Momentum, F⋅∆t = m⋅∆v

A force of 30N is applied to a 15kg block for 8 seconds, so calculate the final speed of the block; F⋅∆t = m⋅∆v, F(t_{f}-t_{0}) = m(v_{f}-v_{0}), F⋅t_{f} = m⋅v_{f}, v_{f} = F⋅t_{f}/m = 30N⋅8s/15kg = 16m/s; F = m⋅a, a = F/m = 30N/15kg = 2m/s^{2}, V_{f} = V_{0}+a⋅t = 0+(2m/s)⋅8s = 16s

A 10kg block moving east at 15m/s strikes a 20kg block initially at rest, so calculate the final speed of the two blocks if they stick together; during a collision, momentum is conserved, p_{0} = p_{f}, m_{1}⋅v_{1}+m_{2}⋅v_{2} = (m_{1}+m_{2})⋅v_{f}, v_{f} = (m_{1}⋅v_{1}+m_{2}⋅v_{2})/(m_{1}+m_{2}) = ((10kg⋅15m/s)+0)/(10kg+20kg) = (150kg⋅m/s)/30kg = 5m/s

Torque is the rotational equivalent of linear force; it is also referred to as the moment, moment of force, rotational force or turning effect; torque = (inertia)⋅(angular acceleration), τ = I⋅α; torque is the rotational equivalent of force, inertia is the rotational equivalent of mass, angular acceleration is the rotational equivalent of the linear acceleration; F = m⋅a is the Newton's second law for translational motion, τ = i⋅α is the Newton's second law for rotational motion; when a force causes an object to rotate, the force creates a torque; torque = force⋅radius, τ = F⋅r

Torque = (inertia)⋅(angular acceleration), τ = I⋅α; torque = force⋅radius, τ = F⋅r

A force of 50N is applied to a 10kg solid disk with a radius of 65cm, so calculate the angular acceleration of the disk; torque = force⋅radius, τ = F⋅r = 50N⋅0.65m = 32.5N⋅m; the inertia formula changes according to the shape of the object, in this case for a solid disk is I = (1/2)m⋅r^{2} = (1/2)(10kg)(0.65m)^{2} = 2.1125kg⋅m^{2}; torque = (inertia)⋅(angular acceleration), τ = I⋅α, α = τ/I = (32.5N⋅m)/(2.1125kg⋅m^{2}) = 15.38rad/s^{2}

Angular momentum, also called moment of momentum or rotational momentum, is the rotational equivalent of linear momentum; angular momentum is an important quantity in physics because it is a conserved quantity, the total angular momentum of a closed system remains constant; angular momentum = (moment of inertia)⋅(angular speed), L = I⋅ω; I = r^{2}⋅m; ω = v/r; L = I⋅ω = r^{2}⋅m⋅v/r = r⋅m⋅v, so the angular momentum L = (radius)⋅(linear momentum)

Angular momentum = (moment of inertia)⋅(angular speed), L = I⋅ω; angular momentum = radius⋅(linear momentum), L = r⋅p = r⋅m⋅v

Angular speed = velocity/radius, ω = v/r

Moment of inertia = radius^{2}⋅mass, I = r^{2}⋅m

A merry-go-round moving with angular speed of 1.8rad/s has a rotational inertia of 100kg⋅m^{2}, and a child jumps on it, and the rotational inertia of the child and the merry-go-round is now 110kg⋅m^{2}, so calculate the new angular speed of the child on the merry-go-round; initial angular momentum = final angular momentum, L_{0} = L_{f}, I_{0}⋅ω_{0} = I_{f}⋅ω_{f}, ω_{f} = (I_{0}⋅ω_{0})/I_{f} = ((100kg⋅m^{2})⋅(1.8rad/s))/(110kg⋅m^{2}) = 1.64rad/s

Two blocks are joined by a rope and are hung on a pulley, and m_{2} > m_{1}, so calculate the acceleration of the blocks; F = m⋅a, a = F/m = (m_{2}⋅g-m_{1}⋅g)/(m_{1}+m_{2}) = g(m_{2}-m_{1})/(m_{1}+m_{2}); F_{2} = T-m_{2}⋅g, -m_{2}⋅a = T-m_{2}⋅g, T = m_{2}⋅g-m_{2}⋅a; F_{1} = T-m_{1}⋅g, m_{1}⋅a = T-m_{1}⋅g, T = m_{1}⋅a+m_{1}⋅g; m_{2}⋅g-m_{2}⋅a = m_{1}⋅a+m_{1}⋅g, m_{2}⋅g-m_{1}⋅g = m_{1}⋅a+m_{2}⋅a, a(m_{1}+m_{2}) = g(m_{2}-m_{1}), a = g(m_{2}-m_{1})/(m_{1}+m_{2})