H1255 by pcherukuri

Work & Energy 

One of the most important concepts in physics Alternative approach to mechanics



Many applications M li ti b beyond d mechanics h i Thermodynamics (movement of heat) Quantum mechanics...



Very useful tools You will learn new (sometimes much easier) ways to solve problems

Forms of Energy 

Kinetic Energy of motion Kinetic: motion. A car on the highway has kinetic energy.

We have to remove this energy to stop it. The breaks of a car get HOT! This is an example of turning one form of energy into another (thermal energy).

Energy Conservation 

Energy cannot be destroyed or created. J t changed Just h d ffrom one form f to t another. th



We say energy is conserved! conserved True for any isolated system. i.e. when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same. The energy of the car “alone” is not conserved... » It is reduced by the braking.



Doing g “work work” on an isolated system y will change g its “energy energy”... gy

Definition of Work: Ingredients: Force (F F), displacement (rr) Work, W, of a constant force F acting through a displacement rr is: W = F rr = F rr cos  = Fr rr

“D t P “Dot Product” d t”

F 

rr

D fi iti off Work... Definition W k 

Only the component of F along the displacement is doing work. Example: Train on a track. F



r F cos 

Example Work & Energy 

A box is p pulled up p a rough g ( > 0)) incline byy a rope-pulleyp p y weight arrangement as shown below. How many forces are doing work on the box? (a) 2 (b) 3 (c) 4

Example Solution 

Draw FBD of box:



Consider direction of motion of the box



Any force not perpendicular to the motion will do work:

N does no work (perp. to v) T does positive work f does negative work mg does negative work

3 fforces do work

Example (constant force) 

A force F = 10 N pushes a box across a frictionless floor for a distance x x = 5 m.

x x Work done by F on box : (since F is parallel to x x) WF = Fxx = F x WF = (10 N) x (5 m) = 50 Joules (J)

Units: Force x Distance = Work Newton x [M][L] / [T]2

mks N-m (Joule)

Meter = Joule [L] [M][L]2 / [T]2

cgs Dyne-cm (erg) = 10-77 J

other BTU = 1054 J calorie = 4.184 4 184 J foot-lb = 1.356 J eV= 1.6x10-19 J

Work & Kinetic Energy: 

A force F = 10 N pushes a box across a frictionless floor for a distance x x = 5 m. The speed of the box is v1 before the push and v2 after the push.

v2 F

i x x

Work & Kinetic Energy... 

Since the force F is constant constant, acceleration a will be constant. We have shown that for constant a: v22 - v12 = 2a(x2-x1) = 2ax. 1/ mv 2 - 1/ mv 2 = max multiply by 1/2m: 2 2 2 1 1/ mv 2 - 1/ mv 2 = Fx But F = ma 2 2 2 1

v2 F

i x x

Work & Kinetic Energy... 

So we find that 1/2mv22 - 1/2mv12 = Fx = WF



Define Kinetic Energy K: K = 1/2mv2 K2 - K1 = WF WF = K (Work/kinetic energy theorem) v2

v1 F

i x x

Work/Kinetic Energy Theorem: {Net Net Work done on object} = {change change in kinetic energy of object}

Wnet  K  K 2  K1 



1 1 2 2 mv 2  mv1 2 2

We’ll p prove this for a variable force later.

Example Work & Energy ď Ź

Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. ď > 0) which slows them down to a stop. Which one will go farther before stopping? (a) m1 (b) m2

Example Solution  

The work-energy theorem says that for any object WNET = K In this example the only force that does work is friction (since both N and mg are perpendicular to the block’s motion).

Example Solution   

The work-energy theorem says that for any object WNET = K In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). The net work done to stop the box is - fD = -mgD.  This work “removes” the kinetic energy that the box had:  WNET = K2 - K1 = 0 - K1

Example Solution 



The net work done to stop a box is - fD = -mgD. This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1 This is the same for both boxes (same starting kinetic energy). m2gD2 m1gD1

Since m1 > m2 we can see that

m2D2 m1D1

D2 > D1

m1 D1

m2 D2

A simple application: Work done by gravity on a falling object  

What is the speed of an object after falling a distance H, H assuming it starts at rest? Wg = F r = mg r cos(0) = mgH v0 = 0

Wg = mgH

rr

mg g

Work/Kinetic Energy Theorem: Wg = mgH = 1/2mv2

v  2 ggH

What about multiple forces? Suppose FNET = F1 + F2 and the displacement is rr. The work done by each force is: W1 = F1 rr

W2 = F2  rr

FNET

WTOT = W1 + W2 = F1 rr + F2 rr = (F F1 + F2 ) rr

WTOT = FTOT rr

It’s the total force that matters!!

rr

Comments:



Time interval not relevant Run up the stairs quickly or slowly...same W

Since W = F r 

No work N k iis d done ifif: F = 0 or r = 0 or  = 90o

Comments... W = F r 

No work done if  = 90o. T

No work done by T.

No work done by N.

Example Work & Energy ď Ź

An inclined plane is accelerating with constant acceleration a. a A box resting on the plane is held in place by static friction. How many forces are doing work on the block?

(a) ( )1

(b) 2

( )3 (c)

Example Solution ď Ź

First, draw all the forces in the system:

FS a

mg g

Example Solution 

Recall that W = F ∆rr so only forces that have a componentt along l the th di direction ti off th the di displacement l t are doing work.

FS a

mg g



The answer is (b) 2. 2

Work by variable force in 33-D: 

Work dWF of a force F acting through an infinitesimal displacement r is:

F r

dW = F r 

The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements:



WTOT = F r

Problem: Block Sliding with Friction...     

Using WNC = K + U As before, U = -mgd WNC = work done by friction = -kmgx. K K = 0 since i th the bl block k starts t t outt and d ends d up att rest. t WNC = U -kmgx = -mgd

x = d / k d

k x

Work done by gravity: 

Wg = F rr = mg rr cos  = -mg y m

mg g

Wg = -mg y

rr 

j y

Depends only on y !

Work done by gravity... 

W NET = W1 + W2 + . . .+ + Wn = F rr 1+ F rr2 + . . . + F rrn = F (rr1 + rr 2+ . . .+ rrn) = F r = F y

m rr1

y y

rr3

Wg = -mg y Depends only on y, y not on path taken!

rrn

r

mg g

rr2

Example

Falling Objects ď Ź

Three objects of mass m begin at height h with velocity 0. One falls straight down down, one slides down a frictionless inclined plane, and one swings on the end of a pendulum. What is the relationship between their velocities when they have fallen to height 0? v=0

v=0

H vf Free Fall (a) Vf > Vi > Vp

vi Frictionless incline (b) Vf > Vp > Vi

vp Pendulum (c) Vf = Vp = Vi

Example

Solution v=0

v=0

H vf Free Fall

Frictionless incline

Pendulum

Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22

v f  v i  v p  2 gH

does not depend on path !!

Lifting a book with your hand: What is the total work done on the book?? 

First calculate the work done by gravity: Wg = mg g r = -mg r



Now find the work done by the hand: WHAND = FHAND r = FHAND r

rr

FHAND v = const a=0 mg g

Example: Lifting a book... Wg = -mg rr WHAND = FHAND rr rr WNET

FHAND v = constt a=0

= WHAND + Wg = FHAND rr - mg rr = (FHAND - mg) rr mg g = 0 since ∆K = 0 (v = const)



So WTOT = 0!!

Textbook

Example: Lifting a book... 

Work/Kinetic Energy Theorem says: W = K {Net Net Work done on object} = {change change in kinetic energy of object} In this case, v is constant so K = 0 and so W must be 0, as we found.

rr

FHAND v = const a=0 mg g

Work done by Variable Force: (1D) 

When the force was constant constant, we wrote W = F x area under F vs. x plot:

F Wg x

x 

For variable force, we find the area by integrating: dW = F(x) dx. F(x) x2

W   F ( x )dx x1

Work/Kinetic Energy Theorem for a Variable Force x2

W   F dx x1 x2

m

dv dx dt

mv v1

F  ma  m dv dt dv dx dv dv = = v dx (chain rule) dt dt dx

dv dx dx

 m  v dv v1

1 1 1  m (v22 v12 )  m v22  m v12  ∆KE 2 2 2

1-D Variable Force Example: Spring ď Ź

For a spring we know that Fx = -kx. -kx F(x) ( )

x2 x

relaxed position

-kx F = - k x1 F = - k x2

Spring... ď Ź

The work done by y the spring p g Ws during g a displacement p from x1 to x2 is the area under the F(x) vs x plot between x1 and x2. ( ) F(x)

x2 x

relaxed position

Ws -kx

Spring... 

The work done by y the spring p g Ws during g a displacement p from x1 to x2 is the area under the F(x) vs x plot between x1 and x2. x2

F(x)

Ws   F ( x )dx x1

x Ws -kx

  ( kx )dx x1

1   kx 2 2 Ws  

x2 x1

1 k x22  x12  2

Example

Work & Energy 

A box sliding g on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest. If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ? (a)

x2  x1

(b) x2  2 x1 x

(c)

x2  2 x1

Example

Solution 

Again, g , use the fact that WNET = K. WNET = WSPRING = -1/2 kx2 K = -1/2 mv2

In this case, and

kx2 =

mv2

In the case of x1

x1  v1

m1 k

x v

m k

Example

Solution

So if v2 = 2v 2 1 and m2 = m1/2

x 2  2v 1

m1 2 k

 v1

2m1 k

x2  2 x1 v2

Problem: Spring pulls on mass. ď Ź

A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the relaxed position if it slides without friction? m

relaxed position

t t h d position iti ((att rest) t) m stretched d m

after release

v m vr

back at relaxed position

Problem: Spring pulls on mass. 

First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring):

Ws  

1 1 1 k x 22  x12    k 0 2  d 2   kd 2 2 2 2

m stretched position (at rest) d m vr

relaxed position i

Problem: Spring pulls on mass. 

Now find the change in kinetic energy of the mass:

1 1 1 ∆K  mv 22  mv12  mv r2 2 2 2

m stretched position (at rest) d m vr

relaxed position i

Problem: Spring pulls on mass. 

Now use work kinetic-energy theorem: Wnet = WS = K. 1 kd 2  2

1 mv r 2 2

vr  d

k m

m stretched position (at rest) d m vr

relaxed position i

Problem: Spring pulls on mass.  

Now suppose there is a coefficient of friction  between the block and the floor The total work done on the block is now the sum of the work done by the spring WS (same as before) and the work done by friction Wf.

Wf = f ∆r = - mg d rr m stretched position (at rest) d m vr

f = mg

relaxed position i

Problem: Spring pulls on mass. 

Again use Wnet = WS + Wf = K 1 2 W  kd S Wf = -mg d 2

1 2 1 2 kd  mgd d  mv r 2 2

1 2 K  mv r 2

vr 

k 2 d  2 μgd m

rr m stretched position (at rest) d m vr

f = mg

relaxed position i

Work by variable force in 3 3--D: Newton’s Gravitational Force 

Work dWg done on an object by gravity in a displacement dr is given by:

^ dWg = Fg dr = (-GMm / R2 r^) (dR r^ + Rd)

dWg = (-GMm / R2) dR

(since r^ ^ = 0, r^ r^ = 1) dR Fg

d R M

Rd



dr m

Work by variable force in 3 3--D: Newton’s Gravitational Force 

Integrate dWg to find the total work done by gravity in a “big” big displacement:



Wg = dWg = ((-GMm / R2) dR = GMm ((1/R2 - 1/R1) R1

Fg(R2)

R2 Fg(R1) R1 M

Work by variable force in 3 3--D: Newton’s Gravitational Force 

Work done depends only on R1 and R2, not on the path taken. taken

 1 1   Wg  GMm    R2 R1 

R1 M

Example

Work & Energy 

A rock is dropped pp from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. hits RE is the radius of the earth earth. What is K2 / K1? (a)

2 2RE

(b) ( ) (c)

3 2 4 3

Example

Solution 

Since energy gy is conserved,, K = WG.  1 1    WG = GMm  R2 R1 

 1 1    ∆K = c   R2 R1 

Where c = GMm is the same for both rocks 2RE

Example

Solution

 1 1  ∆K = c     R2 R1 

 1 1  1 1   c    2 RE  RE 2RE 



F the For th first fi t rock: k K1 = c 



F the For th second d rock: k K 2 = c 

 1 1  2 1   c    3 RE  RE 3RE 

So:

2 K2 3 4 =  K1 1 3 2

2RE

Newton’s Gravitational Force Near the Earth’s Surface: 

Suppose R1 = RE and R2 = RE + y

 R2  R1   RE  y   RE    GM       y Wg  GMm   GMm m   2   R  y R   R   R1R2     E  E E  GM  but we have learned that  2   g  RE 

So:

Wg = -mgy m

RE+ y

RE M

Conservative Forces: 

We have seen that the work done by gravity does not depend on the path taken. m

 1 1   Wg  GMm   R R  2 1 

R2 R1 M m

Wg = -mgh

Conservative Forces: 

In general general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservative conservative.



 1 1  Gravity is a conservative force: Wg  GMm R  R   2 1 



Gravity near the Earth’s surface: Wg  mgy



A spring produces a conservative force: Ws  

1 k x 22  x12  2

Conservative Forces: ď Ź

We have seen that the work done by a conservative force does not depend on the path taken taken. W2 W1 = W2

ď Ź

Therefore the work done in a closed path is 0. WNET = W1 - W2 = W1 - W1 = 0

W1 W2

Example

Conservative Forces ď Ź

The p pictures below show force vectors at different p points in space for two forces. Which one is conservative ? (a) 1

(b) 2

y x

(1)

(2)

Example

Solution ď Ź

Consider the work done by force when moving along different paths th iin each h case: WA = WB

(1)

WA > WB

(2)

Example 

In fact, you could make money on type (2) if it ever existed: Work done by this force in a “round round trip trip” is > 0! Free kinetic energy!! WNET = 10 J = K

W=0

W = 15 J

Note: NO REAL FORCES OF THIS TYPE EXIST, S , SO FAR AS WE KNOW

W = -5 J

W=0

Potential Energy 

For any conservative force F we can define a potential energy function U in the following way: W =

U  F.drdr = -U

 The

work done by a conservative force is equal and opposite to the change in the potential energy function function. r2



This can be written as:

F.drr r2

U = U2 - U1 = -W = -

Gravitational Potential Energy 

We have seen that the work done by gravity near the Earth’s Earth s surface when an object of mass m is lifted a distance y is Wg = -mg y



The change in potential energy of this object is therefore: U = -Wg = mg y

y

Wg = -mg y

Gravitational Potential Energy 

So we see that the change in U near the Earth Earth’s s surface is: U = -Wg = mg y = mg(y2 -y1).



So U = mg y + U0 where U0 is an arbitrary constant constant.



Having an arbitrary constant U0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. j

y2 y1

Wg = -mg y

Potential Energy Recap: 

For any conservative force we can define a potential energy function U such that: S2

U = U2 - U1 = -W = -

F.drr S1



The potential energy function U is always defined only up to an additive constant constant. can choose the location where U = 0 to be anywhere h convenient. i t

 You

Conservative Forces & Potential Energies (stuff you should know): Work W(1-2)

Force F ^

Fg = -mg j Fg = 

GMm ^ r 2 R

Fs = -kx k

-mg(y ( 2-y1)

 1 1   GM  GMm   R R  2 1 



1 k x 22  x12  2

Change in P.E U = U2 - U1 mg(y ( 2-y1)

P.E. function U mgy + C

 1 GMm 1    C  GMm GM    R  R2 R1 

1 k x22  x12  2

1 2 kx  C 2

(R is the centercenter-toto-center distance, x is the spring stretch)

Example

Potential Energy ď Ź

All springs p g and masses are identical. ((Gravity y acts down). ) ď&#x192;§Which of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position?

(a) 1 (b) 2 (c) same

(1)

(2)

Example

Solution ď Ź

The displacement p of ((1)) from equilibrium q will be half of that of ((2)) (each spring exerts half of the force needed to balance mg)

0 d 2d (1)

(2)

Example

Solution 1 The potential energy stored in (1) is 2  kd2  kd2 2

The potential energy stored in (2) is

1 2 k2d  2kd2 2

The spring P.E. is twice as big in (2) !

0 d 2d (1)

(2)

Conservation of Energy 

If only conservative forces are present, the total kinetic plus potential energy of a system is conserved, i.e. the total “mechanical energy” is conserved. (note: E=Emechanical throughout this discussion) E=K+U E = K + U  using K = W = W + U W W) = 0  using U = -W = W + ((-W)

E = K + U is constant constant!!! !!!  

Both K and U can change, but E = K + U remains constant. But we’ll see that if non-conservative forces act then energy can be b di dissipated i t d iinto t other th modes d (th (thermal,sound) l d)

Example: The simple pendulum 

Suppose we release a mass m from rest a distance h1 above its lowest possible point.  What is the maximum speed of the mass and where pp does this happen?  To what height h2 does it rise on the other side?

h2 v

Example: The simple pendulum ď Ź ď Ź

Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) Choose y = 0 at the bottom of the swing, y choice)) and U = 0 at y = 0 ((arbitrary E = 1/2mv2 + mgy y

y=0

h2 v

Example: The simple pendulum 

E = 1/2mv2 + mgy. mgy  Initially, y = h1 and v = 0, so E = mgh1.  Since E = mgh1 initially, E = mgh1 always since energy is conserved. conserved

y=0

Example: The simple pendulum  1/2mv2 

will be maximum at the bottom of the swing swing. 1/ mv2 = mgh So at y = 0 v2 = 2gh1 2 1 v 

2 gh h1

y = h1 y=0

h1 v

Example: The simple pendulum ď Ź ď Ź

Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum height on the other side will be at y = h1 = h2 and v = 0. The ball returns to its original height.

y = h1 = h2 y=0

Example: The simple pendulum ď Ź

The ball will oscillate back and forth forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1/2mv2 + mgy = K + U = constant. t t

Example: The simple pendulum ď Ź

We can also solve this by choosing y = 0 to be at the original position of the mass, and U = 0 at y = 0. E = 1/2mv2 + mgy.

y=0 h1

h2 v

Example: The simple pendulum 

E = 1/2mv2 + mgy. mgy  Initially, y = 0 and v = 0, so E = 0.  Since E = 0 initially, E = 0 always since energy is conserved.

y=0

Example: The simple pendulum  1/2mv2 

will be maximum at the bottom of the swing swing. 1/ mv2 = mgh So at y = -h1 v2 = 2gh1 2 1

v 

2 gh1

y Same as before! y=0 y = -h h1

h1 v

Example: The simple pendulum ď Ź ď Ź

Since 1/2mv2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0. The ball returns to its original height.

y=0 Same as before!

Example: Airtrack & Glider ď Ź

A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ?

m d

Example: Airtrack & Glider  

Kinetic+potential energy is conserved since all forces are conservative. Choose initial configuration to have U=0. K = -U

1  m  M v 2  mgd 2

m d

Problem: Hotwheel ď Ź

A toy car slides on the frictionless track shown below below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2. ď&#x192;§ Find v1 and v2.

v2 d v1

Problem: Hotwheel...   

K+U energy is conserved, conserved so E = 0 K = - U Moving down a distance d, U = -mgd, K = 1/2mv12 Solving for the speed: v1  2 gd

d v1

Problem: Hotwheel...   

At the end end, we are a distance d - h below our starting point point. U = -mg(d - h), K = 1/2mv22 Solving for the speed: v 2  2 g d  h 

d-h d

v2 h

Non--conservative Forces: Non 

If the work done does not depend on the path taken taken, the force is said to be conservative.



If the work done does depend on the path taken, taken the force is said to be non-conservative.



An example of a non non-conservative conservative force is friction friction. When pushing a box across the floor, the amount of work that is done by friction depends on the path taken. » Work W k done d iis proportional ti l tto th the llength th off th the path! th!

Non--conservative Forces: Friction Non 

Suppose you are pushing a box across a flat floor floor. The mass of the box is m and the coefficient of kinetic friction is k.



The work done in pushing it a distance D is given by: Wf = Ff • D = -kmgD. D

Ff = - kmg

Non--conservative Forces: Friction Non



Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just Wf = -mgL. Clearly, the work done depends on the path taken.



Wpath 2 > Wpath 1



path 1 p path 2 A

Generalized Work/Energy Theorem: 

Suppose FNET = FC + FNC (sum of conservative and nonconservative forces).



The total work done is: WNET = WC + WNC



The Work/Kinetic Energy theorem says that: WNET = K.  WNET = WC + WNC = K  WNC = K - WC



But WC = -U So

WNC = K + U = Emechanical h i l

Generalized Work/Energy Theorem: WNC = K + U = Emechanical 

The change in kinetic+potential energy of a system is equal t the to th workk done d on it b by non-conservative ti fforces. Emechanical =K+U of system not conserved! Iff all the forces f are conservative, we know that K+U energy is conserved: K + U = Emechanical = 0 which says that WNC = 0, which makes sense. If some non-conservative force (like friction, a “push” or a “pull”) does work, K+U energy will not be conserved and WNC = E, which also makes sense sense.

Problem: Block Sliding with Friction 

A block slides down a frictionless ramp ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k.  How far, x, does the block go along the bottom portion of the track before stopping?

k x

Problem: Block Sliding with Friction...     

Using WNC = K + U As before, U = -mgd WNC = work done by friction = -kmgx. K K = 0 since i th the bl block k starts t t outt and d ends d up att rest. t WNC = U -kmgx = -mgd

x = d / k d

k x