• Wrong document? Swap it for free
  • Written by students who passed
  • Immediately available after payment
  • Read online or as PDF
Sell
Where do you study
Your language
Document preview thumbnail
Preview 1 out of 2 pages
Class notes

Physics Forces Notes and Practice Problems

Document preview thumbnail
Preview 1 out of 2 pages

Comprehensive notes covering forces and Newton's Laws of Motion. Includes force diagrams, net force calculations, friction (static and kinetic), tension, normal force, gravity, inclined plane problems, and step-by-step example calculations. These notes are organized, easy to follow, and include practice problems with explanations to help students prepare for quizzes, exams, and homework assignments. Ideal for introductory physics students looking for a clear understanding of force concepts and problem-solving strategies.

Content preview

Section 12 1 .




Newtons law of Universal Gravitation ;




Fgrav .
= Em , me


r2

↳ distance between their centers
Nm2
G = 6 67. x 10- "
1g2
this force acts between every pair of objects
this force is always


Gravity :



we know Eg =
-may
Only vertical displacement gives work
·



nonzero

if we let
lay) = h , then IWg) mgh :


-

moving down Eg and by are parallel ,
and Eg does positive work :
Wg togh =




When an object is lifted or lowered ,
the applied force does work too .

·

if
object starts &-
ends at rest , then work from applied Force is equal in magnitude , opposite in


sign , to from .
work
gravity


Section 5 5 .




if the force varies as position changes ,
work is given by
W= S"Fx x)dx
Xy
for 2D 3D motion we
or
get
·


,




Way =
F dr .




A



·

this integral follows from A to i
path taken point point


Section 5 . 6 :




recall that Esp = -KX (if Xo is equilibrium position
if from
you hang a mass a spring, gravity has effect of
shifting equilibrium position , so we can still

consider just the spring.
* kx (Fsp)


>
work done by a
spring is
Wsp kx
-
=
=




neg cause spring is trying
it is to
get back to equilibrium K 63 5 N
=
1450 NIm
·


. =




EX : Solved Problem 5 2 .
0 0435m
.




F 63 5N
=
.


,
1X = 0 0435m
.



,
M= 0 .

075kg ,
Vo = Omis k =

+muz
1450 : E (0 875) v .





V =

Document information

Uploaded on
June 17, 2026
Number of pages
2
Written in
2025/2026
Type
Class notes
Professor(s)
Unknown
Contains
All classes
$6.19

Wrong document? Swap it for free Within 14 days of purchase and before downloading, you can choose a different document. You can simply spend the amount again.
Written by students who passed
Immediately available after payment
Read online or as PDF

Sold
0
Followers
0
Items
14
Last sold
-




Why students choose Stuvia

Created by fellow students, verified by reviews

Quality you can trust: written by students who passed their tests and reviewed by others who've used these notes.

Didn't get what you expected? Choose another document

No worries! You can instantly pick a different document that better fits what you're looking for.

Pay as you like, start learning right away

No subscription, no commitments. Pay the way you're used to via credit card and download your PDF document instantly.

Student with book image

“Bought, downloaded, and aced it. It really can be that simple.”

Alisha Student

Working on your references?

Create accurate citations in APA, MLA and Harvard with our free citation generator.

Working on your references?

Frequently asked questions