Jacobsthoughtsblog's Blog

Work and Energy

10 Jun 2010 1 Comment

by Jacob in Physics Tags: physics

Answers to last post‘s problems:

1. 200 N
2. 10 N on a 5 kg mass produces an acceleration of 2m/s².  Integrating twice with respect to t with t being two gives us a travel distance of 4 meters.
3. The force on the larger box is 10g-8g, and the force on the smaller box is 8g-10g, giving the smaller box a net force of 2g upwards.  This simplifies to 19.62N, and an acceleration of 2.4525m/s².  Integrating once over 3 seconds gives a velocity of 7.3575 m/s.
4. The units of k are N/m=(kg m/s²)/m=kg/s².  Isolating s gives s²=kg/k, or s=√(kg/k) so the period is C√(mass/k) where C is a multiplier constant.

Work (in one dimension) is defined as the distance traveled by an object times the component in the direction of movement of the force on this object.  It can be stated as W=Fcos(θ)Δx or W=F(x)*Δx.  Force has two commonly used units.  One, the Joule, J for short, is defined as N*m.  The other, the electron volt, eV, is .00000000000000000016J.  If the force is changing, or is in more than one dimension, work is defined as W=∫F ds, where s is the direction of the force, and the distance that force is used over.

If we substitute Newton’s second law into the work equation we obtain W=m*a*Δx.  Since v²=v²(intial)+2aΔx then 1/2(v²-v²(intial))=aΔx, and substituting in we get W=1/2m*v²-1/2m*v²(intial).  1/2mv² is known as kinetic energy.  That means that work done on a system is the change of kinetic energy in that system.  K is the letter we use to designate kinetic energy.

There is also the potential for an object to have energy or potential energy.  The final kinetic energy of an object is equal to the initial potential energy when under conservative forces.  A conservative force is one that is used when the particle under force travels a closed path.  When v(initial)=0, then ½v²=aΔx, or 1/2mv²=maΔx.  The most common usage of potential energy is to find the velocity of an object falling from a height(h), where a=g, and Δx=h.  Potential energy can also be used to solve spring equations.  U is the letter used to designate potential energy an can also be defined as, W=-U.

When an object has balanced forces working on it it is said to be in equilibrium.  There are three types of equilibrium:

1. Stable equilibrium:  This means that small displacements will result in a force towards point of equilibrium (i.e. a spring).
2. Unstable equilibrium: This means that small displacements will result in a force away from the equilibrium position (i.e. pushing something off a table).
3. Neutral equilibrium: This means that small displacements will result in no force at all (i.e. sliding something across a flat surface).

Energy is always conserved.  This means that the energy that enters a system is equal to the energy that leaves.  This energy can come in different forms such as kinetic energy, potential energy, heat, molecular bonds, and mass of which the conversion is defined by the equation E=mc², where E is the energy that the mass would produce and c is the speed of light.

There are certain cases in which the laws of energy need to be refined.  One of them is at velocities close to that of light.  At these velocities kinetic energy is defined as K=½Ev²/c² where E is the rest energy of the mass at hand.  This translates into K=½mv² at low speeds.

The other case is when dealing with light, electron orbits, and matter waves. We describe energy as quantized meaning put in discrete amounts.  Electrons can only absorb energy in quantized amounts due to the fact that it is only allowed certain orbits.  The energy absorbed is E=hf where f is the frequency and h is Planck’s constant equal to 6.6 *10^-34 J*s.

Practice problems:

1. What is the work used to accelerate a box weighing 4 kg from 10m/s to 15m/s over the course of .1 seconds?
2. If a 60 kg object is dropped from 100 meters up, what is the final velocity and kinetic energy of the object right before it hits the ground?
3. If there is a spring with spring constant k, what is the potential energy of a particle attached to its spring displaced by x meters from equilibrium.

Newton’s three laws

16 May 2010 1 Comment

by Jacob in Physics Tags: physics

The answers to last weeks problems:

1. 50km/h at 36.9° north of east.  This can be obtained by using the Pythagorean theorem and by taking the arc-tangent of 3/4.
2. The ball travels 10 meters in the square root of 2 seconds.  You can derive this by finding the x and y components, finding when the ball hits the ground, and putting that into an equation to find x distance traveled.  (x=vt)
3. Substitute x=vt into y=vt – ½gt².  This yields y(x)=v(y)*(x/v(x)) – ½g(x/v(x))²=(v(y)/v(x))*x – (g/2v²(x))*x².  This becomes y(x)=(tanθ)*x – (g/2v²cos²θ)*x².  This is the path of a projectile.  To find the time, T, at the end of the objects flight one must set the y equation to zero, making y=v(y)t - ½gt²=0 → t*(v(y) - ½gt) =0.  This means flight time is T=2v(y)/g = 2v/g *sinθ.  The total distance traveled, R, is defined as R=v(x)=(v cosθ)*(2v/g *sinθ)=2v²/g *sinθ*cosθ.  Because 2*sinθ*cosθ=sin(2θ).  This means that R=v²/g *sin(2θ).

There are three basic laws of linear motion in classical mechanics for solid objects being,

1. An object in rest stays in rest unless acted upon by a force.  An object in linear motion stays with that velocity unless acted upon by a force.
2. The force exerted on an object produces an acceleration proportional to the mass of an object, when not acted upon by other forces.  This is know as F=ma (the true equation is F=dp/dt, but that’s for later.)
3. Every force has counteracting forces that sum to be equal and in the opposite direction.

These are known as Newton’s three laws.  The units for force are in Newtons, defined as 1N(Newton)=1kg*m/s², or the force it takes to accelerate 1 kilogram 1 meter per second squared.    The force of gravity is mass times the gravitational acceleration meaning that the weight of 1 kilogram near the Earth’s surface is 9.81 Newtons.

There are several types of forces, the for fundamental ones being the gravitational force, the electromagnetic force, the strong nuclear force and the weak nuclear force.  The force experienced as two objects push against each other is classified under the electromagnetic force, but in classical mechanics is usually called a contact force.  The contact force perpendicular to a surface is called a normal force, and is equal to the gravitational force as shown by Newton’s third law.  The force experienced parallel to a surface is called frictional.  We will not be dealing with frictional forces in this post.  Another force is that made by a spring.  The equation for how forces are experienced by the object on the spring is F=-kΔx, with k being a constant based on the spring, and the negative sign meaning that the force is in the opposite direction of how much the object is moved from the position with no force.  This equation is very useful in oscillations and quantum mechanics.

A trick for solving forces problems is to use what is called a free body diagram.  This consists of drawing plotting a point for the particle under force, choosing a convenient coordinate system, drawing the forces using arrows, labeling the unknowns, and writing out the equation.  This method also works for a system of several particles, drawing out a free body diagram for each of the particles.

Now for practice questions:

1. A mass of 0.5kg is attached to a spring with a spring constant of 200N/m.  This mass is displaced by 1 meter.  What is the force exerted on the mass?  Assume no friction, and that the system is on a table.
2. A 5kg box is pushed by a force of 10 Newtons, for a duration of 2 seconds.  How far has the box moved?
3. Two masses are attached to each other on a pully system.  If one box is 10 kg, and the other is 8 kg, what is the velocity of the smaller box after three seconds?
4. Using the spring equation, also known as Hook’s equation, find out the period of a mass oscillating on a spring, omitting any multiplier constants.

Velocity in two dimensions, and projectiles

14 Apr 2010 Leave a Comment

by Jacob in Physics Tags: physics

(prerequisite includes trigonometry) Let’s start off with the problems from last lecture.

1. The acceleration is about 1o meters per second squared.  This can be obtained through the formula d^2 (cx^t)/dx^2=d (ctx^t-1)/dx =ct(t-1)x.
2. 6 seconds passed, and is going 30 meters a second.  This can be obtained by dimensional analysis.
3. It starts to fall at 1 second, hits the ground after 2 seconds, and is going 10 meters per second.  Notice that it ends at the same velocity at which it begins.
4. I will work out the last problem.  Δx=v(average)t=1/2 (v(first) +v)t= 1/2 (v(first) +v)v-v(first)/a=v²-v(first)²/2a.  This simplifies to (if v (first) equals zero), v²=2aΔx

As we all know, not all movement is in strait lines, but can change in multiple directions.  For example, if you want to find your velocity in east-west/ north-south components then you would have to use the Pythagorean theorem and trigonometric functions.  It is velocity*cos(Θ) is the x component, and velocity*sin(Θ) is the y component where Θ is the counter clockwise angle from the x axis.

This can also be used for projectiles.  The x position is the x component times the amount of time the projectile was moving, and the y position is the y component times time, minus 1/2 gt².

Practice problems:

1. Find the velocity of a car moving with components of 30km/h north, and 40km/h east.
2. A golf ball is hit at a 45 degree angle at 10m/s.  Disregarding air resistance, how far does the ball go, and how much time does it take to get there?
3. Make an equation for how far an object travels.

Instantaneous velocity, and constant acceleration.

29 Mar 2010 Leave a Comment

by Jacob in Physics Tags: physics

Note: Due to the upcoming holiday, next weeks lecture will also be on Monday. (pre-requisites include high school algebra and differential calculus.)

To start off, lets review last week‘s questions I gave you.

1. The average velocity is two meters per second (2m/s).  You get this answer by dividing the 10 meters by 5 seconds.
2. The time taken is 2% of an hour, or an equivalent amount.  This can be obtained by dividing 2 kilometers by 100 km/s.
3. The average velocity is 666 and 2/3 meters, or an equivalent amount.  This is a much trickier problem, which can be obtained by converting the velocity to meters per second (22 and 2/9) and multiplying by the time.

As you all know, things don’t stay at one speed but change speed.  This change is called acceleration. It is in units of distance per time squared, x/t².  The “t²” means that it is the change of velocity over a certain amount of time.  You can obtain the average acceleration by dividing the average velocity by the time, or dv/dt=a.  This becomes the second derivative of the distance time curve, or d²x/dt².  Using differential calculus we can work out a)v=v(first)+at, and b)dx=at².

Now for the tricky part.  What happens if you want to find the velocity of an object at a single point and the velocity is changing throughout the time interval.  This is called instantaneous velocity.  To find this we must take the first derivative of  position with respect to time.

Practice problems.

1. The acceleration at which objects fall (if air resistance is negligible) is about 5t² meters.  What is acceleration due to gravity?
2. A man throws an apple off a 500 meter cliff.  Using the gravitational acceleration you worked out, how much time passed when the apple falls 180 meters, and how fast is it going?
3. Someone throws a graduation cap into the air, at a velocity of 10 meters a second.  Using the gravitational acceleration you worked out, how long till the cap starts to fall, how long till the cap reaches the ground, and what is its speed as it falls?
4. (Extera credit: Using last lecture’s “average acceleration equation,” and this lecture’s “acceleration a equation,” compute v in terms of x, v(first), and a)

Basic velocity

24 Mar 2010 1 Comment

by Jacob in Physics Tags: physics

All objects exist in four dimensions.  There are three of space (up and down, north and south, and east and west) and one of time.  In one spatial dimension, of which I will be referring to for mathematical simplicity, position is referred to as “x”.  x can be height or distance from a single point.  Time is usually expressed in seconds, and is referred to as “t”.  Velocity is the change of position over time, or the derivative of position with respect to time, or the slope of a function on the position versus time graph.  It is expressed as distance per time (i.e. meters per second) and is referred to as “v”.  To make simple enough for high school algebra, I will use only average velocity for this lesson.  For linear velocities (constant ones), the velocity can be obtained by dividing the change in position (the final position minus the first one), divided by the change in time (the final time minus the first). As an equation, Δx/Δt=v, or v average, where Δ(uppercase delta) means “the change of”.  Average velocity is also the final velocity minus the initial velocity, divided by two (v average=1/2 v(last)-v(first)).  (We will always regard v(first) as 0.  Also, we will not use this in this section, but refer back to it for next lecture.)

Here are some practice problems:

1. An object starts off at point 0 and time 0.  In 5 seconds it moves 10 meters.  What is its average velocity?

2. A train goes 2km at 100km per hour.  How long did the train take to travel that distance?

3. A car moves 80 kilometers per hour for 30 seconds.  How many meters did the car move in that time?

Answer in the comments section, the answers will be gone over next week in the next lecture.