PHYS16 – Lecture 15

Work and Energy October 13, 2010

Agenda

• Administration– Homework for Week 5– Exam

• What have we learned so far? What do we still need to know?

• Energy• Mechanical Work

Description of Motion – What else do we need?

• We have:– Laws of Calculus – Displacement, Velocity and

Acceleration– Newton’s Laws – F=ma – Concept of Momentum

We need energy…

Definition of Energy• Energy– A quantity whose expenditure or transformation allows for

physical activity– An ability to drive motion– A capacity for action

• Scalar Quantity• Unit = Joule (J) = kg·m2/s2

• Comes in many forms– Thermal– Chemical– Mechanical!!!!!

Mechanical Energy• Kinetic Energy (K)– energy stored in the movement of an

object

• Potential Energy (U) – energy stored in the configuration of a system– Gravitational Potential Energy

– Spring Potential Energy

2

21 mvK

mghU

2

21 kxU

Energy can be transformed

• Wyle E. Coyote• http://www.youtube.com/watch?v=Jnj8mc04r9E&feature=related

Practice Question

• A 0.50 kg vase falls from 3.0 m. What is the kinetic energy of the vase just before it hits the ground?

A) 0 JB) 15 JC) 1.5 JD) 2.3 J

Practice Question

• A 0.50 kg vase falls from 3.0 m. What is the potential energy of the vase before it falls?

A) 0 JB) 15 JC) 1.5 JD) 2.3 J

We need work…

Definition of Work

• Mechanical Work (W) – energy transferred to an object due to the action of a force(+) transfer to object(-) transfer from object

xdFW

Aside on Dot Product

• Dot Product is one way to multiply two vectors– Basically just multiply components and add– Dot Product is a scalar

...

,...),(

,...),(

2211

21

21

BABABA

BBB

AAA

A

B

Aside on Dot Product

• Dot Product is one way to multiply two vectors– Basically just multiply components and add– Dot Product is a scalar

A

B181230

)2,5(

)6,6(

BA

B

A

Aside on Dot Product

• Dot Product is one way to multiply two vectors– Basically just multiply components and add– Dot Product is a scalar– Or multiply magnitudes and cosine angle between

the vectors

18)56cos(296)cos(

)2,5(

)6,6(

BABA

B

A

A

B

θ=56°

Work with a Constant Force

• Force = Constant, then can take force outside integral

cosxFW

xFW

Work with a Variable Force

• Force = Constant, then can take force outside integral

dxFW x

F x

x

Practice Question

• I pull a 4.0 kg sled a distance of 5.0 m. I pull the sled using a rope at a 30.0 degree angle with a force of 5.0 N. What is the work done by me?

A) 0 JB) 20 JC) 25 JD) 22 J

Practice Question

• A force is given by Fx = 3x2+2. What is the work done by the force for moving an object from x=0.0 m to x=4.0 m?

A) 72 JB) 50 JC) 0 JD) 200 J

Work – Energy Theorem

• Work = the transfer of Energy• Energy = the ability to do work

EW

Work done byExternal Force

Change in Energyto the system

Work and grav. potential energy• If I lift an object, how much work did I do on the object?• Use work-energy theorem to derive gravitational potential

energy

Uyymg

Udymg

UxdF

UEW

y

y

)(

)(

0

0

Force and displacementare both downward

Work and spring potential energy• If mass on a spring moves, how much work is done by spring?• Use work-energy theorem to derive spring potential energy

Uxxk

Udxkx

UxdF

UEW

x

x

)(21

)(

20

2

0

Work done by system is negative

Force and displacementare in opposite directions

Work and Kinetic energy• If an object speeds up, how much work is done on object?• Use work-energy theorem to derive kinetic energy

)(21 so...

1

)(21

)(21)(

20

2

20

2

20

2

vvmΔK

madtdvmv

vF

vvmdxdt

dtdF

vvmdxFdxd

KxdF

KEW

Assume K=mv2/2

and prove left side =right side

Just multiply and divideby dt since dt/dt=1

Now take derivative andremember to use chainrule