×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

## PhysforSci&EngrLecI

by: Collin Turner

19

0

3

# PhysforSci&EngrLecI PHYS013

Marketplace > Howard University > Physics 2 > PHYS013 > PhysforSci EngrLecI
Collin Turner
HU
GPA 3.6

Staff

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Physics 2

This 3 page Class Notes was uploaded by Collin Turner on Monday October 12, 2015. The Class Notes belongs to PHYS013 at Howard University taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/222154/phys013-howard-university in Physics 2 at Howard University.

×

## Reviews for PhysforSci&EngrLecI

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/12/15
Physics for Scientists and Engineers last updated 9221 PHYS 013 Lecture notes and addenda T Hubsch 1 The differential equation a body falling under the in uence of gravity and resisted by air drag dvt dt may easily be solved by the trial and adjust7 method First we note that the equation is linear in vt This allows us to solve the equation by means of a superposition addition of the solutions of the two simpler equations digit g and 111215 7005 2 a b The first of these Eq 2a is solved by 0115 010 gt as seen before while Eq 2b is solved by remembering that a derivative of the exponential function is proportional to the exponential function itself so that 0215 Ugoe btm This indeed satisfies Eq 2b as is easy to verify by substituting mg7bvt or d2 97005 1 Then we look for the general solution to the original equation 1 in the form vt ABtOeDt 3 From this we calculate the derivative to be C1005 Dt W 7 B ODe 4 Substituting 4 in the left hand side and 3 in the right hand side of Eq 1 we obtain b b b b 13 ODeDt g7 7A 1325 Gem g7 7A i 7327 705D m m m m Equating the coef cients of t0 l 007L311 t1 t and am we have that b b b Bg77A 0773 and OD 77C 5abc m m m Of these Eq 5b forces B 0 after which Eq 5a sets A while Eq 50 sets D 7 This then produces vt Oe btm which indeed has one integration constant C as should be the case for a rst order differential equation To determine this last constant we can use the initial condition noticing that m d f 00Tg0 e 110 is the initial at t 0 speed and so can write 0 UO 7 whereupon vt E 110 7 btW 17 e btm vo e btm b b b Finally we know that the body has reached the terminal speed 01 if it no longer accelerates 26 when the gravitational force is cancelled by the air drag mg va so that UT Indeed when t 7 00 then e btm 7 0 and so voo 11 This provides an auxiliary check for our general solution Notice that the value of the terminal speed UT is independent of the initial speed vo Also note that vt does not in fact reach UT at any nite time but merely approaches it asymptotically Handout 3 Newton s Laws Dr A Rahman 1 1 Newton s Laws Newton7s Laws form the foundation of motion in this course They form a link between the kinematic and dynamic motion of bodies You will nd that a sound understanding of Newton7s laws will serve as the basic insight needed to solve any problem 11 Description We shall provide more detailed elaboration on each law in the following sections Let7s begin by stating Newton7s three laws 0 N1 If no force acts on a body7 then the body7s velocity cannot change that is7 the body cannot accelerate Restated If no net force acts on a body Fnet 7 then the body7s velocity cannot change that is7 the body cannot accelerate 0 N2 The net force sum of all the forces on a body is equal to the product of the body7s mass and the acceleration of the body Restated llet m6 or EllE m6 0 N3 When two bodies interact7 the force of body one on body two is equal and opposite to the force of body two on body one Restated F12 7F21 111 N1 Inertia Newton7s rst law is sometimes called the law of inertia The meaning of this physical law is summed up in the common adage 77A body at rest tends to stay at rest until placed in motion and a body in motion tends to stay in motion until a force is acted upon it77 This means that a body will continue its motion along said path as long as no force acts on it 112 N2 2115 mg Newton7s second law relates dynamics forces to kinematics motion It is said that just about every classical mechanics equation can be derived from it Please keep this in mind as you work with it A couple of points regarding the nature of force as presented in this law First7 as you can see from the equation7 the force is actually a net force7 ie the sum of all forces acting on the body Second7 force is a vector7 mass is a scalar7 and acceleration is a vector That means solving problems often requires resolving the force vector into its components These components represent the forces components in a certain direction 113 N3 ActionReaction Newton7s third law simply states that for every action force acting on an object there is an equal an opposite action force exerted from the object This law gives rise to the term action reaction pairs of forces An example of this is the normal force of contact that is the reaction force to placing a mass on a surface They are equal in magnitude but opposite in direction Handout 3 Newton s Laws Dr A Rahman 2 12 Friction Friction is a non conservative force that serves to reduce a particle7s velocity There are many types of friction air resistance drag static and kinetic friction 121 Properties of Friction There are two types of friction that will be studied static and kinetic friction Static friction exists between a static object and the surface it rests on The amount of force required to move the object to get it moving is the force required to overcome static friction Kinetic friction is the friction that exists between a moving object and the contact surface Here are some fundamental properties of friction that will be used o If the body does not move then the static friction force f and the component of F are equal in magnitude and f5 is directed opposite to F o The magnitude of has a maximum value of jam that is given by fZ max MEN where us is the coef cient of static friction and N is the magnitude of the normal force on the body from the surface Note that if the magnitude of the component of F that is parallel to the surface exceeds lm then the body begins to slide along the surface If the body begins to slide along the surface the magnitude of the frictional force rapidly decreases to a value of given by ukN where Mk is the coef cient of kinetic friction and N is the magnitude of the normal force on the body from the surface During sliding the frictional force opposes the motion of the sliding body

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Allison Fischer University of Alabama

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over \$600 per month. I LOVE StudySoup!"

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com