SlideShare a Scribd company logo
1 of 26
Download to read offline
Solving Collinear
Dynamics Problems
       CP Physics
       Elise Burns
The most important thing
is...
     You must show supporting work!
     You must show supporting work!
     You must show supporting work!
     You must show supporting work!
     You must show supporting work!
We’ll do an example...

 Here’s a typical problem: A rope exerts a
 280 N force while pulling an 80 kg water
 skier. A 110 N friction force opposes the
 skier’s motion. If the skier starts at rest,
        determine his acceleration.
 We’ll go through each step, page by page.
STEP 1: Draw a picture.




A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 2: Label all known information




T = 280-N
m = 80-kg
f = 110-N
  vi = 0

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 3: Label the unknown information




T = 280-N
m = 80-kg
f = 110-N
  vi = 0
  a = ???

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 4: Draw a Motion Diagram.

                                          v   v       v


                                                  a




T = 280-N
m = 80-kg
f = 110-N
  vi = 0
  a = ???

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 5: Draw a Free Body Diagram. (Don’t forget the axis!)

                                            v       v           v       +y


                                                        a
                                                                             +x




T = 280-N                                                   N


m = 80-kg                                                           T
                                                f
f = 110-N                                                   w

  vi = 0
  a = ???

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 6: WRITE OUT THE COMPONENTS OF NEWTON’S
  SECOND LAW.
                                            v       v           v

                                                                        +y
                                                        a


                                                                             +x
                                                            N
T = 280-N
m = 80-kg                                       f
                                                                    T


f = 110-N     Σ Fx   = max   Σ Fy   = may                   w


  vi = 0
  a = ???

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 7: Substitute in the specific forces for “Net Force”.
                                               v   v           v

                                                                       +y
                                                       a


                                                                            +x
                                                           N
T = 280-N
m = 80-kg                      f
                                                                   T


f = 110-N ΣFx = max ΣFy = may                              w


  vi = 0 T - f = max N - w = may
  a = ???

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 8: Plug in all known information into the equation.

                                           v       v           v

                                                                       +y
                                                       a


                                                                            +x
                                                           N
T = 280-N
m = 80-kg                                      f
                                                                   T


f = 110-N ΣFx = max                                        w


  vi = 0 T - f = max                               Σ Fy= may
  a = ??? 280 - 110 = (80)ax                       N - w = may

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 9: If necessary, solve for weight... This is not needed here.

                                                v       v           v

                                                                            +y
                                                            a


                                                                                 +x
                                                                N
T = 280-N
m = 80-kg                                           f
                                                                        T


f = 110-N ΣFx = max                                             w


  vi = 0 T - f = max                                    Σ Fy= may
  a = ??? 280 - 110 = (80)ax                            N - w = may

A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
STEP 10: Solve the problem. Put units on your answer!

T = 280-N                                    v       v           v


m = 80-kg                                                a
                                                                         +y




f = 110-N                                                    N
                                                                              +x




  vi = 0                                                             T
  a = ???      Σ Fx = max                        f

             T - f = max                        w

       280 - 110 = (80)ax                     ΣFy = may
               170 = (80)ax
                                              N - w = may
         2.1 m/s2 = ax
A rope exerts a 280 N force while pulling an 80 kg water
skier. A 110 N friction force opposes the skier’s motion.
  If the skier starts at rest, determine his acceleration.
Question: How did I know whether to
use the x or y axis equation? They both
have acceleration in the equation.
   There are several hints:
   1. The forces given (T and f) are both in
   the x-axis equation.

   2. The object is not moving vertically, so
   that ay is equal to zero anyway.

   3. You aren’t given Normal force, so you
   can’t solve the y-axis equation.
Question: How did I know whether the
person or the boat was the system?
 There are several hints:

 1. The mass of the person is given and the
 mass of the boat was not.

 2. The question asks for the acceleration of
 the person.

 3. The friction force opposes the person’s
 motion.
Question: Could I have drawn a sketch where
 the person moved from right to left, instead of
 left to right?

                   YES!

      You would get the same answer.
          The same exact answer!

(The sign may be different if you don’t also
change your +x axis to go in the direction of
                motion.)
Question: Could I have used an axis
that showed the +x pointing opposite
the direction of motion?

                    YES!

       You would get the same answer.

The sign would be opposite, but as long as you
    show your work, that’s not a problem.
Question: Do I really have to write every
 single step out? It’s so annoying. Why are
 you being so picky about this?
YES! You absolutely must show all of your work.
I have 3 reasons.

1. Science is about solving problems. Solving problems is a
process. I am checking to make sure that you learn the process.

2. I cannot read your mind. The only way you can communicate
your understanding of this material is by putting pen to paper
and showing me.

3. If you want partial credit when you err, you must show all of
your steps so I can give you points for what you did correctly.
Want to see another
   solved problem?

A 60-kg person falls from a storage building and lands
stiff-legged in some mud. The person stops with an
acceleration of 490 m/s2, after sinking into the mud.
Determine the average force of the mud in stopping the
person.
A 60-kg person falls from a storage building and lands stiff-legged in some
mud. The person stops with an acceleration of 490 m/s2, after sinking into
the mud. Determine the average force of the mud in stopping the person.

 m = 60-kg                            V          N          +y


a = 490 m/s2                           V
                                           a
                                                                 +x


   vf = 0                              V         w

  Nmud = ?? ΣFx = max                          Σ Fy
                                               = may
                                         N - w = may
                                       N - mg = may
   Notice that in this problem,
   you must solve for weight,     N - (60)(10) = (60)(490)
         using w = mg.                 N - 600 = 29,400
                                         N = 30,000 N
Want to see one more
   solved problem?

1000-kg out of control car initially moving at 20 m/s
enters a horizontal escape surface made of loose
gravel. The force of friction is 2500-N. Determine
the acceleration of the car while coming to a stop.
1000-kg out of control car initially moving at 20 m/s enters a
horizontal escape surface made of loose gravel. The force of friction
is 2500-N. Determine the acceleration of the car while coming to a
stop.                                               N                +y



m = 1000-kg                                                               +x



f = 2500-N                                   f

   vf = 0                                          w     v       v    v




vi = 20 m/s
                                                             a




                          Σ Fx
                           = max                  Σ Fy= may
  a = ???
                       - f = max                  N - w = may
                   - 2500 = (1000)ax
                 -2.5 m/s2 = ax
Referring to this same situation, how far does the car travel while
stopping? How would you approach this?
                                                                      +y
                                                     N

m = 1000-kg                                                                +x



f = 2500-N                                    f

   vf = 0                                           w     v       v    v




vi = 20 m/s
                                                              a




   a = ???                Σ Fx
                           = max                   Σ Fy= may
                       - f = max                   N - w = may
                   - 2500 = (1000)ax
                 -2.5 m/s2 = ax
Use kinematics to find out how far the car travels while stopping.
 You’ve got initial and final velocities given, plus you just solved for
 acceleration. All that’s left to do is pick the equation.             +y
                                                      N

 m = 1000-kg                                                                    +x



 f = 2500-N                                     f

    vf = 0                                            w    v       v        v



 vi = 20 m/s                                                   a




-2.5 m/s2 = ax
                      vf2 = vi2 + 2ad    You are not
                   02 = (20)2 + 2(-2.5)d allowed to
                      0 = 400 +(-5)d
                        -400 =(-5)d         forget
                          80m = d        kinematics!
Now it’s your turn...

Solve the 6 problems on the worksheet.

SHOW ALL STEPS!

Don’t skip any.

Use the list on the notes page to assist you!
Other Physics News...
Complete Dynamics Worksheet #1 - 6

Inspect Lab Journals... Post-Lab Quiz on this week’s
lab will be on Monday!

Test on Tuesday on Newton’s First and Second Laws.

You may do the Newton’s Laws Internet Activity for
Extra Credit. It must be handed in by Tuesday.

Project Analysis should be complete up through Step J.

More Related Content

Viewers also liked

Physics On the Road-Lesson 12
Physics On the Road-Lesson 12Physics On the Road-Lesson 12
Physics On the Road-Lesson 12
cstraughan
 
Physics Chapter 3 And 5
Physics Chapter 3 And 5Physics Chapter 3 And 5
Physics Chapter 3 And 5
guest10e136
 
Physics problem 35
Physics problem 35Physics problem 35
Physics problem 35
bribri2321
 
Newton’s second law problems solving strategies 12 march 2013(2)
Newton’s second law problems solving strategies 12 march 2013(2)Newton’s second law problems solving strategies 12 march 2013(2)
Newton’s second law problems solving strategies 12 march 2013(2)
Raymond Ngobeni
 
Newton’s Laws Of Motion
Newton’s Laws Of MotionNewton’s Laws Of Motion
Newton’s Laws Of Motion
eliseb
 
AP Physics - Chapter 6 Powerpoint
AP Physics - Chapter 6 PowerpointAP Physics - Chapter 6 Powerpoint
AP Physics - Chapter 6 Powerpoint
Mrreynon
 

Viewers also liked (11)

Physics On the Road-Lesson 12
Physics On the Road-Lesson 12Physics On the Road-Lesson 12
Physics On the Road-Lesson 12
 
Physics Chapter 3 And 5
Physics Chapter 3 And 5Physics Chapter 3 And 5
Physics Chapter 3 And 5
 
Physics problem 35
Physics problem 35Physics problem 35
Physics problem 35
 
Physics formulas list
Physics formulas listPhysics formulas list
Physics formulas list
 
Newton’s second law problems solving strategies 12 march 2013(2)
Newton’s second law problems solving strategies 12 march 2013(2)Newton’s second law problems solving strategies 12 march 2013(2)
Newton’s second law problems solving strategies 12 march 2013(2)
 
Newton’s Laws Of Motion
Newton’s Laws Of MotionNewton’s Laws Of Motion
Newton’s Laws Of Motion
 
Problems and solutions statistical physics 1
Problems and solutions   statistical physics 1Problems and solutions   statistical physics 1
Problems and solutions statistical physics 1
 
AP Physics - Chapter 6 Powerpoint
AP Physics - Chapter 6 PowerpointAP Physics - Chapter 6 Powerpoint
AP Physics - Chapter 6 Powerpoint
 
Physics formula list 2
Physics formula list 2Physics formula list 2
Physics formula list 2
 
Physics Numerical Problem, Metric Physics, Kinematic Problems Karachi, Federa...
Physics Numerical Problem, Metric Physics, Kinematic Problems Karachi, Federa...Physics Numerical Problem, Metric Physics, Kinematic Problems Karachi, Federa...
Physics Numerical Problem, Metric Physics, Kinematic Problems Karachi, Federa...
 
AP Physics - Chapter 4 Powerpoint
AP Physics - Chapter 4 PowerpointAP Physics - Chapter 4 Powerpoint
AP Physics - Chapter 4 Powerpoint
 

More from eliseb (15)

Lenses And Mirrors
Lenses And MirrorsLenses And Mirrors
Lenses And Mirrors
 
Reflection And Refraction
Reflection And RefractionReflection And Refraction
Reflection And Refraction
 
Light
LightLight
Light
 
Sound
SoundSound
Sound
 
Waves Ppp
Waves PppWaves Ppp
Waves Ppp
 
Conservation Of Momentum
Conservation Of MomentumConservation Of Momentum
Conservation Of Momentum
 
Projectiles
ProjectilesProjectiles
Projectiles
 
Vectors
VectorsVectors
Vectors
 
What Is Physics
What Is PhysicsWhat Is Physics
What Is Physics
 
Keplers Laws
Keplers LawsKeplers Laws
Keplers Laws
 
Transformers - Movie Physics
Transformers - Movie PhysicsTransformers - Movie Physics
Transformers - Movie Physics
 
Coulomb's Law
Coulomb's LawCoulomb's Law
Coulomb's Law
 
Bernoulli's Principle
Bernoulli's PrincipleBernoulli's Principle
Bernoulli's Principle
 
Universal Gravitation PPP
Universal Gravitation PPPUniversal Gravitation PPP
Universal Gravitation PPP
 
Circular Motion PPP
Circular Motion PPPCircular Motion PPP
Circular Motion PPP
 

Recently uploaded

The basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptxThe basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptx
heathfieldcps1
 
ppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyesppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyes
ashishpaul799
 
Liberal & Redical Feminism presentation.pptx
Liberal & Redical Feminism presentation.pptxLiberal & Redical Feminism presentation.pptx
Liberal & Redical Feminism presentation.pptx
Rizwan Abbas
 

Recently uploaded (20)

An Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptxAn Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptx
 
Championnat de France de Tennis de table/
Championnat de France de Tennis de table/Championnat de France de Tennis de table/
Championnat de France de Tennis de table/
 
Open Educational Resources Primer PowerPoint
Open Educational Resources Primer PowerPointOpen Educational Resources Primer PowerPoint
Open Educational Resources Primer PowerPoint
 
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
 
The Benefits and Challenges of Open Educational Resources
The Benefits and Challenges of Open Educational ResourcesThe Benefits and Challenges of Open Educational Resources
The Benefits and Challenges of Open Educational Resources
 
B.ed spl. HI pdusu exam paper-2023-24.pdf
B.ed spl. HI pdusu exam paper-2023-24.pdfB.ed spl. HI pdusu exam paper-2023-24.pdf
B.ed spl. HI pdusu exam paper-2023-24.pdf
 
Mbaye_Astou.Education Civica_Human Rights.pptx
Mbaye_Astou.Education Civica_Human Rights.pptxMbaye_Astou.Education Civica_Human Rights.pptx
Mbaye_Astou.Education Civica_Human Rights.pptx
 
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
Incoming and Outgoing Shipments in 2 STEPS Using Odoo 17
 
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
50 ĐỀ LUYỆN THI IOE LỚP 9 - NĂM HỌC 2022-2023 (CÓ LINK HÌNH, FILE AUDIO VÀ ĐÁ...
 
slides CapTechTalks Webinar May 2024 Alexander Perry.pptx
slides CapTechTalks Webinar May 2024 Alexander Perry.pptxslides CapTechTalks Webinar May 2024 Alexander Perry.pptx
slides CapTechTalks Webinar May 2024 Alexander Perry.pptx
 
The basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptxThe basics of sentences session 4pptx.pptx
The basics of sentences session 4pptx.pptx
 
Morse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptxMorse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptx
 
Salient features of Environment protection Act 1986.pptx
Salient features of Environment protection Act 1986.pptxSalient features of Environment protection Act 1986.pptx
Salient features of Environment protection Act 1986.pptx
 
Application of Matrices in real life. Presentation on application of matrices
Application of Matrices in real life. Presentation on application of matricesApplication of Matrices in real life. Presentation on application of matrices
Application of Matrices in real life. Presentation on application of matrices
 
Gyanartha SciBizTech Quiz slideshare.pptx
Gyanartha SciBizTech Quiz slideshare.pptxGyanartha SciBizTech Quiz slideshare.pptx
Gyanartha SciBizTech Quiz slideshare.pptx
 
Word Stress rules esl .pptx
Word Stress rules esl               .pptxWord Stress rules esl               .pptx
Word Stress rules esl .pptx
 
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdfTelling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
Telling Your Story_ Simple Steps to Build Your Nonprofit's Brand Webinar.pdf
 
ppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyesppt your views.ppt your views of your college in your eyes
ppt your views.ppt your views of your college in your eyes
 
Liberal & Redical Feminism presentation.pptx
Liberal & Redical Feminism presentation.pptxLiberal & Redical Feminism presentation.pptx
Liberal & Redical Feminism presentation.pptx
 
2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptx2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptx
 

Dyn Prob Slvg Notes

  • 1. Solving Collinear Dynamics Problems CP Physics Elise Burns
  • 2. The most important thing is... You must show supporting work! You must show supporting work! You must show supporting work! You must show supporting work! You must show supporting work!
  • 3. We’ll do an example... Here’s a typical problem: A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration. We’ll go through each step, page by page.
  • 4. STEP 1: Draw a picture. A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 5. STEP 2: Label all known information T = 280-N m = 80-kg f = 110-N vi = 0 A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 6. STEP 3: Label the unknown information T = 280-N m = 80-kg f = 110-N vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 7. STEP 4: Draw a Motion Diagram. v v v a T = 280-N m = 80-kg f = 110-N vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 8. STEP 5: Draw a Free Body Diagram. (Don’t forget the axis!) v v v +y a +x T = 280-N N m = 80-kg T f f = 110-N w vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 9. STEP 6: WRITE OUT THE COMPONENTS OF NEWTON’S SECOND LAW. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N Σ Fx = max Σ Fy = may w vi = 0 a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 10. STEP 7: Substitute in the specific forces for “Net Force”. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max ΣFy = may w vi = 0 T - f = max N - w = may a = ??? A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 11. STEP 8: Plug in all known information into the equation. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max w vi = 0 T - f = max Σ Fy= may a = ??? 280 - 110 = (80)ax N - w = may A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 12. STEP 9: If necessary, solve for weight... This is not needed here. v v v +y a +x N T = 280-N m = 80-kg f T f = 110-N ΣFx = max w vi = 0 T - f = max Σ Fy= may a = ??? 280 - 110 = (80)ax N - w = may A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 13. STEP 10: Solve the problem. Put units on your answer! T = 280-N v v v m = 80-kg a +y f = 110-N N +x vi = 0 T a = ??? Σ Fx = max f T - f = max w 280 - 110 = (80)ax ΣFy = may 170 = (80)ax N - w = may 2.1 m/s2 = ax A rope exerts a 280 N force while pulling an 80 kg water skier. A 110 N friction force opposes the skier’s motion. If the skier starts at rest, determine his acceleration.
  • 14. Question: How did I know whether to use the x or y axis equation? They both have acceleration in the equation. There are several hints: 1. The forces given (T and f) are both in the x-axis equation. 2. The object is not moving vertically, so that ay is equal to zero anyway. 3. You aren’t given Normal force, so you can’t solve the y-axis equation.
  • 15. Question: How did I know whether the person or the boat was the system? There are several hints: 1. The mass of the person is given and the mass of the boat was not. 2. The question asks for the acceleration of the person. 3. The friction force opposes the person’s motion.
  • 16. Question: Could I have drawn a sketch where the person moved from right to left, instead of left to right? YES! You would get the same answer. The same exact answer! (The sign may be different if you don’t also change your +x axis to go in the direction of motion.)
  • 17. Question: Could I have used an axis that showed the +x pointing opposite the direction of motion? YES! You would get the same answer. The sign would be opposite, but as long as you show your work, that’s not a problem.
  • 18. Question: Do I really have to write every single step out? It’s so annoying. Why are you being so picky about this? YES! You absolutely must show all of your work. I have 3 reasons. 1. Science is about solving problems. Solving problems is a process. I am checking to make sure that you learn the process. 2. I cannot read your mind. The only way you can communicate your understanding of this material is by putting pen to paper and showing me. 3. If you want partial credit when you err, you must show all of your steps so I can give you points for what you did correctly.
  • 19. Want to see another solved problem? A 60-kg person falls from a storage building and lands stiff-legged in some mud. The person stops with an acceleration of 490 m/s2, after sinking into the mud. Determine the average force of the mud in stopping the person.
  • 20. A 60-kg person falls from a storage building and lands stiff-legged in some mud. The person stops with an acceleration of 490 m/s2, after sinking into the mud. Determine the average force of the mud in stopping the person. m = 60-kg V N +y a = 490 m/s2 V a +x vf = 0 V w Nmud = ?? ΣFx = max Σ Fy = may N - w = may N - mg = may Notice that in this problem, you must solve for weight, N - (60)(10) = (60)(490) using w = mg. N - 600 = 29,400 N = 30,000 N
  • 21. Want to see one more solved problem? 1000-kg out of control car initially moving at 20 m/s enters a horizontal escape surface made of loose gravel. The force of friction is 2500-N. Determine the acceleration of the car while coming to a stop.
  • 22. 1000-kg out of control car initially moving at 20 m/s enters a horizontal escape surface made of loose gravel. The force of friction is 2500-N. Determine the acceleration of the car while coming to a stop. N +y m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a Σ Fx = max Σ Fy= may a = ??? - f = max N - w = may - 2500 = (1000)ax -2.5 m/s2 = ax
  • 23. Referring to this same situation, how far does the car travel while stopping? How would you approach this? +y N m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a a = ??? Σ Fx = max Σ Fy= may - f = max N - w = may - 2500 = (1000)ax -2.5 m/s2 = ax
  • 24. Use kinematics to find out how far the car travels while stopping. You’ve got initial and final velocities given, plus you just solved for acceleration. All that’s left to do is pick the equation. +y N m = 1000-kg +x f = 2500-N f vf = 0 w v v v vi = 20 m/s a -2.5 m/s2 = ax vf2 = vi2 + 2ad You are not 02 = (20)2 + 2(-2.5)d allowed to 0 = 400 +(-5)d -400 =(-5)d forget 80m = d kinematics!
  • 25. Now it’s your turn... Solve the 6 problems on the worksheet. SHOW ALL STEPS! Don’t skip any. Use the list on the notes page to assist you!
  • 26. Other Physics News... Complete Dynamics Worksheet #1 - 6 Inspect Lab Journals... Post-Lab Quiz on this week’s lab will be on Monday! Test on Tuesday on Newton’s First and Second Laws. You may do the Newton’s Laws Internet Activity for Extra Credit. It must be handed in by Tuesday. Project Analysis should be complete up through Step J.