Physics Chapter 5

Work and Energy

WorkWork - (if force is constant) – is the product of the force exerted on an object and the distance the object moves in the direction of the force.W = F||d

Work is a scalar quantity and can be positive or negative, but you don’t have to indicate direction/angle in your answer.

Unit is Joule (N•m)The object must MOVE in the DIRECTION

(positive or negative) that the force is exerted on THE OBJECT.Holding a flag in the air (no work on the flag)Walking forward while holding the flag in the

air (no work on flag)

Work and Direction of Force

If force is applied at an angle, then the work must be computed using the component of force in the direction of the motion.Ex: you are pushing a lawnmower…

Work and Direction of Force

Force (parallel to ground) doing work is FcosWork (W) = Fcosd= FdcosWork that the grass/lawn is doing on the

mower = Ffrictiond

F

dFf

Fcos

ExampleHow much work is done on a

vacuum cleaner pulled 4.0 meters by a force of 45 N at an angle of 30.0 degrees above horizontal?

Here’s the equation: W = Fd = F cos d = Fdcos

W = Fd = F cos d = 45Ncos30X 4.0 m = 156 Joules

30

45N

Fcos

D

mg

mgx

Wgrav = mg sin d

Or Wgrav = mgd if lifted vertically

Gravity ExampleHow heavy is a barrel that is

lifted 45 meters vertically using 670 joules of work?

Formula for work done against gravity.

Wagainst grav = -Wgrav =-mgd670J = -mgdm=-670J/(gd) = -670/(-9.81m/s2X45m) = 1.5 kg

Practice 5A page 170 and Section Review on

page 171

Happy Thursday!

Please take out 5A and section review for me to check.

Energy - an object has energy if it can produce a change in itself or its surroundings.

Kinetic Energy - energy due to the motion of an object.

KE = 1/2mv2

Kinetic Energy

Doing Work to Change Kinetic Energy

Kinetic Energy is related to the amount of work done on the object.Work done equals the KE

gained by the object.

Example pg. 173

A 7.0 kg bowling ball moves at 3.00m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g ping pong ball move in order to have the same kinetic energy as the bowling ball?

Known, unknowns and formulas

mbb=7.0kg

Vbb=3.00m/s

mppb=2.45g = 0.00245kg

KEbb = 1/2mbbvbb2 = KEppb

= 1/2mppbvppb2

Solve for Kinetic Energy of Bowling Ball

KEbb = 1/2mbbvbb2 = (½(7.0kg)

(3.00m/s)2=

=31.5 J = 32J

Solve for vppb

KEbb = 32J = KEppb = 1/2mppbvppb2

vppb = 32J/(1/2mppb)

vppb = 32J/[(1/2)(0.00245kg)]

vppb =160m/s

Example 2

What is the speed of a 0.20 kg baseball that has a Kinetic Energy of 390Joules?

m= 0.20 kg, KE = 390J, v=?KE = ½ mv2 v= √2KE/mv= √2(390J/0.20kg) = 62m/s

Example 3What is the mass of a

satellite traveling at 92 km/h with a kinetic energy

of 240,000 J

KE = 240,000, v=92km/h?????92km/hX1000/3600=26m/sKE=1/2mv2 , m=2KE/v2 m=2(240,000J)/(26m/s)2 = 710kg

5B page 174

Effect of Doing Work

As positive work is done on an object (lifting a box), energy is transferred to that object (it can now fall and crush something).

As negative work is done on an object (lowering a box), energy is transferred from the object to the one lowering the box.

The Work-Energy TheoremThe net work done on an

object is equal to its change in kinetic energy.Wnet = KEf - KEi = KE

If the net work is positive, the kinetic energy increases.

If the net work is negative, the kinetic energy decreases.

Example

On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.20 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?

Known, unknowns and formulasms=10.0kg

Vs=2.20m/sk = 0.100KEf = 0Fnet = Fk…the only force working

once the sled is kicked, but it is NEGATIVE in direction!!

Fnet=-Fk=-k Fn = k mgWnet = -Fkd = KEf – KEi

-k mg d = -KEi = -1/2msvs2

Solve for d

d =-1/2msvs2 /-(k mg)=

-1/2(10.0kg)(2.20m/s)2/-(0.10)(10.0kg)(9.81m/s2)

=2.46mIt still travels a positive distance, but it

slows down as it is traveling so its change in kinetic energy, friction and acceleration are negative.

Practice 5C

Potential Energy - energy stored in an object because of its state or position.

Gravitational potential energy depends on an object’s position above a zero level.PEg = mghPEg = mghf - mghi = Work done

Potential energy can be stored in bent or stretched objects=Elastic Potential Energy

PEelastic=1/2kx2

x = distance compressed or stretchedk = spring constant

g is positive for potential energy!

What is the Potential Energy of a 24.0 kg mass 7.5 m above the ground

PEg = mgh =

(24.0)(9.81)(7.5m) = 1.8X103 J

What is the Potential Energy of a spring

stretched 5.0 cm that has a spring constant of 79

N/m?PEE = ½ kx2 =

½ (79N/m)(.05m)2 = = 0.099J

Total Potential energy =Elastic Potential Energy + Gravitational Potential Energy

PETotal=1/2kx2 + mgh

ExampleA 70.0 kg stuntman is attached to

a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?

Known, unknows and formulasms=70.0kg

hi=50.0m, hf=?...hf= 50m-44m=6.0mLi= 15.0mLf = 44.0mL = 29.0m =xk=71.8N/mPETotal=1/2kx2 + mghf

PETotal=1/2(71.8N/m)(29m)2+ (70.0kg)(9.81m/s2)(6.0m) =

3.43X104J

Swing problem…

hyp adj

Height above zero point = hyp-adj

(hyp)cos = adj

swing

Problems 5D, page 180 and section review

page 181

Potential and Kinetic Energy

When a ball is thrown into the air, the kinetic energy you give the ball is transferred to potential energy and then back into the same amount of kinetic energy -- Total Energy is constant.ETOTAL = KET + PET

Mechanical Energy

Mechanical Energy (ME) is the sum of KE and ALL forms of PE associated with an object or group of objects.

ME = KE + PE

Law of Conservation of Energy

States that within a closed, isolated system, energy can change form, but the total amount of energy is constant.(Energy can be neither created nor destroyed).

MEi=MEf

KEi + PEg,i+ PEE,i = KEf + PEg,f+ PEE,f

½ mvi2+mghi + ½ kxi

2= ½ mvf

2+mghf+½kxf2

Example 1

Starting from rest, a child slides down a frictionless slide from an initial height of 3.00 m. What is his speed at the bottom of the slide? Assume he has a mass of 25 kg.

hi = 3.00m, hf = 0.00m

Vi=0

m=25.0kgg=9.81m/s2

Vf=?

Use the Conservation of Mechanical Energy Formula

(no PEE)KEi + PEg,i = KEf + PEg,f

mghi = ½ mvf2

vf=(ghi )/(½ )vf= (9.81)(3.00m)/(1/2) =7.67m/s

Example 2. A brick with a mass of 2.1 kg falls from a height of 2.0 m and lands on a relaxed spring with a constant of 3.5 N/m. How far will the spring be compressed? Assume final position is 0.0 m and final KE is 0.

Given:

Find:

Formula:

Solution:

m = 2.1 kg hi = 2.0 m g = 9.81 m/s2 k = 3.5 N/m

x

k mgh , PE P PE 2fg,iE,, 2

1ifE, xsoEPE ig

m 4.9 N/m 3.5

m) )(2.0m/s kg)(9.81 2(2.1

k

2mgh

2i

x

No KE initial or final

Practice 5E

Conservation of Energy Lab

Power

Work is not affected by the time it takes to perform it.

Power is the rate of doing work.P=Wnet/t = Fnet,||d/t OR P= Fnet,|| v

Remember… Wnet = KEF = maFriction/Resistance is a force that needs to be included in Fnet calculations

Units are in watts (W)One Watt = 1 Joule/secondKilowatt = 1000 watts

Power

Example 1. How long does it take for a 1.70 kW steam engine to do 5.6 x 106 J of work? (assume 100% efficiency.)

Given:

Find:

Formula:

Solution:

P = 1.70 x 103 W W = 5.6 x 106 J

t

t

W P

s 10 x 3.3 W10 x 1.70

J 10 x 5.6

P

W t 3

3

6

Power

Example 2. How much power must a crane’s motor deliver to lift a 350 kg crate to a height of 12.0 m in 4.5 s?

Given:

Find:

Formula:

Solution:

m = 350 kg h = 12.0 m t = 4.5 s g = 9.81 m/s2

P

t

mgh

t

dF P

net

W9200 s 4.5

m) )(12.0m/s kg)(9.81 (350

t

mgh P

2

P=Wnet/t = Fnetd/t

Power

Example 3. A 1.2 x 103 kg elevator carries a maximum load of 900.0 kg. A constant frictional force of 5.0 x 103 N retards the elevator’s upward motion. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?

Hint: The key to this problem is to find the required net force required in the vertical

Fnet = Fw elevator + Fw load + Ffriction = g(me + ml) + Ff

Fnet = 9.81 m/s2(1.2 x 103 kg + 900.0 kg) + 5.0 x 103 N

Fnet = 25 601 N (without rounding)

Given:

Find:

Formula:

Solution:

Fnet = 25 601 N = 3.00 m/s

P

netF P

W000 77 m/s) N)(3.00 601 (25 F P net

Example 4

A 1450 kg car accelerates from rest to 5.60m/s in 5.00sec. If Fk is 560.0N, what is the power developed by the engine?

Given: m=1450kg, vi=0, vf=5.60m/s, t=5.00s, Fk=560.0N

Need: acceleration to determine net force (F=ma)! Then you need x (this is d) to find Work. Then you can calculate Power.

a=v/ t=5.60m/s/(5.00s)=1.12m/s2

**engine has to overcome friction(Fk) and cause acceleration (F=ma)

Fnet=Fk+ma=560.0N+(1450kg)X(1.12m/s2) = 2184N

**use an old formula to find xx=1/2(vi+vf)Xt= ½(0+5.60m/s)X5s = 14m

**Now find powerP=W/ t = Fnetd/ t = 2184NX14m/5.00sP=6115Watts = 6.115kW

Practice 5F

Review problems in class

1,7,10,12,13,17,19,22,24,25,30, 33,35