lecture 34gut/phys_2514/links/lect_34.pdfa 10 kg box slides 4:0 m down the frictionless ramp shown...

Physics 2514Lecture 34

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

Physics 2514 – p. 1/13

Information

Information needed for the examExam will be in the same format as the practice with thesame number of questionsBring a # 2 pencil & eraserCalculators will be allowedNo cell phones, no laptops, . . .Only exam, pencil, eraser, calculator allowed on desk.Bring student id with you

You will need to knowStudent id numberDiscussion section #Your name

Physics 2514 – p. 2/13

Material to be covered

This exam will cover chapters 9, 10, & 11Impulse-momentum theorem

~J = ∆~p with ~J =∫ tf

ti

~F dt and ~p = m~v

Third law ⇒ momentum conservation ~pi = ~pf if noexternal forces

Work-Kinetic energy theoremWnet = ∆K with Wnet =

∑i Wi, and Wi =

∫ sf

si

~Fi · d~s

Conservation of mechanical energy E = T + U

Potential energy Wi = −∆U

Physics 2514 – p. 3/13

Review Momentum

Momentum given by ~p = m~v

Newton’s second law defines forces in terms of momentumchange

d~p

dt= ~Fnet ⇒ ∆~p = ~J =

∫ tf

ti

~Fnet dt

(~J is the impulse)

Newton’s third law leads to momentum conservation~F12 = −~F21 leads to m1~v1i + m2~v2i = m1~v1f + m2~v2f

This implies that no external forces are acting on the twoobjects

Physics 2514 – p. 4/13

Review Momentum

Object released from rest near the Earth’s surface

~J = ∆~p ⇒ −mg∆t = mvyf

⇒ vyf = −g∆t

0 = MVyf + mvyf

Physics 2514 – p. 5/13

Mechanical Energy Review

Kinetic energy K = 12mv2 energy due to motion, is always

positive

Potential energy gravitational U = mgy; spring U = 12k(∆s)2

Stored energy ∆K = −∆U

Total mechanical energy E = Ki + Ui constant assuming nofrictional forces, kinetic & potential energies are mechanicalenergy

Energy is a scalar (not a vector)

Zero of potential energy is arbitrary you decide where to setit Only potential energy differences matter

Energy units are kg-m2/s2 = JoulesPhysics 2514 – p. 6/13

Review of Work

Introduced concept of work:Energy added due to forces acting on an object,Wnet =

∑i Wi = ∆K;

Work for an individual force given by Wi =∫ sf

si

~Fi · d~s andif conservative Wi = −∆U

If the work done is independent of the path, the force isconservative

Force can be written as a potential (true for gravity andspring forces), The mechanical energy is conserved.

Friction is not a conservative forceThe work done depends on the path;

Force from potential energy Fs = −dUds

Physics 2514 – p. 7/13

Steps in Problem Solving

Steps in problem solving1.) Rewrite the problem eliminating all extraneous

information. (What are you given, what are you looking,what are the constraints);

2.) Draw a diagram along with a coordinate system, labeleach object with the variables associated with it (includeforces, initial, final momentum, initial and final energy,and work and impulse where appropriate);

3.) What are the known and unknown quantities, whichunknowns are you solving for;

4.) Write down the equations associated with the problem,and solve the problem algebraically

5.) Finally, substitute numbers into the equation, andcalculate the numerical solution

Physics 2514 – p. 8/13

Example

The skiing duo of Brian (80 kg) and Ashley (50 kg) is always a crowdpleaser. In one routine, Brian starts at the top of a 200 m long 20◦

slope. Ashley waits for him halfway down. As he skis past, she leapsinto his arms and carries her the rest of the way down. What is theirspeed at the bottom of the slope (assume there is no friction)?An object (80 kg) starting from rest slides 100 m down a 20◦ frictionlessincline. It collides inelastically with a second object (50 kg), which is atrest, and can continues for an additional 100 m. What is its final speed?

0

1

2

PSfrag replacements

m = 80 kg

m = 130 kg

m = 130 kg

mgs

n

v, a

θ

Physics 2514 – p. 9/13

Example

An object (m1 = 80 kg) starting from rest slides 100 m down a θ = 20◦

frictionless incline. It collides inelastically with a second object(m2 = 50 kg), which is at rest, and can continues for an additional100 m. What is its final speed?

0

1

2

PSfrag replacements

m = 80 kg

m = 130 kgs

mg

n

v, a

θ

Speed at 100 m:m1gy2 = 1

2m1v2

1+ m1gy1

yn = sn sin θ

v2

1= 2g sin θ ∆s, v1 = 25.9 m/s

∆s = s2 − s1

Inelastic collision ∆v

m1v1 = (m1 + m2)v′

1, v′

1= 15.9 m/s

Speed at 200 m (s = 0, y = 0)(m1 + m2)gy1 + 1

2(m1 + m2)v′2

1= 1

2(m1 + m2)v2

0

v2

0= v′2

1+ 2gs1 sin θ, v = 30.4 m/s

Physics 2514 – p. 10/13

Example Energy Conservation

A 10 kg box slides 4.0 m down the frictionless ramp shown in thefigure. It then collides with a spring whose spring constant is250 N/m.

1. What is the maximum compression of the spring?2. At what compression of the spring does the box have the

maximum velocity?

Physics 2514 – p. 11/13

Example Energy Conservation

What is the maximum compression of the spring (m = 10 kg,k = 250 N/m)?

1. Select coordinate system that simplifies the problem. (Selected todivide the problem into gravity only, and gravity plus spring)

PSfrag replacementss

x

y

Use energy conservation (K = 0 initialand final positions)

Initial energy:Ei = mgsi sin(30) = 196 J

Final energy:Ef = mgsf sin(30) + 1

2ks2

f

Energy conservation: Ei = Ef

⇒ mgsi sin(30) = mgsf sin(30) + 1

2ks2

f

⇒ sf = −1.46 mPhysics 2514 – p. 12/13

Example Energy Conservation

At what compression of the spring does the box have themaximum velocity?

Mass in contact w/spring

PSfrag replacements

mg

−ks

nNet force at initial contact along s:

Fs = −mg sin(30)

Net force at maximum displacement:Fs = −ks − mg sin(30) > 0 (s < 0)

Net Force vs Position

0-0.7-1.4

-500

100

200

0-0.7-1.4

-500

100

200

PSfrag replacements

Fs

s

Max speed (kinetic energy) at Fs = 0

Fs = 0 = −ks − mg sin(30)

⇒ s = −mg

ksin(30) = −0.196

Physics 2514 – p. 13/13