In physics, work is done by forces on an object over a distance.

1a. W = F d (Joule, J)

Force is parallel to distanceForce is at an angle to distance

1b. W = F⋅d⋅Cosθ

2. Wnet = ∆E ext

3. Kinetic Energy, Ek = 1/2m⋅v2 (Joule, J)

4. Potential Energy, EP = m⋅g⋅h (Joule, J)

5. Power, P = W∕∆t (Watts, W)

Chapter 6 Work & EnergyAP

PHYS

5. Spring, EP = ½kx2 (Joule, J)

Fsp

Fsp

Deformation of a Spring

Natural length

Compression

Extension

-x

x = 0

+x

The deformation (x)of a spring is measured relative to itsunstretched length.

Fsp = −k⋅x

Spring Constant(N/m)

Force (N)

Deformation (m)

Draw the diagram below. It illustrates Hooke’s Law.

Warm Up (11/20/15) — Hooke’s LawAP

PHYS

Weight = m⋅g = Fw

Fsp = −k⋅x1

x2x1

Lab 10 — Determining the Spring Constant, kAP

PHYS

AP

PHYS

Chapter 6 Work & Energy

Guiding Questions

▼ How is the energy of a system defined? ▼ How is work represented graphically? ▼ What is mechanical energy and what factors affect its conservation?

AP

PHYS

Warm Up (12/02/15) — Types of Energy

Describe the types of energy present in each photo

burning transforms some of the chemical energy of hydrogen and oxygen into electromagnetic energy (light energy) and thermal energy.

Energy can be either kinetic or potential.

burning transforms some of the chemical energy of hydrogen and oxygen into electromagnetic energy (light energy) and thermal energy.

x2x1

Types of EnergyAP

PHYS

AP

PHYS

Mechanical energy is not always conserved, but total energy is.

Mechanical Energy

The work done by the force of friction as the piston expands and compresses inside the cylinder generates heat.

Conservation of Energy AP

PHYS

A 50. kg sled is pulled for 5.2 m with a horizontal force of 100 N, starting from rest. Find the final speed. Ignore friction.

Work Changes EnergyAP

PHYS

Wnet, ext = ∆E; system: sled, m = 50kg

WFg + WFN + WF = ∆KE; A one body system can only have KE0 0 → because θ = 90°

0 + 0 + FdCosθ = (KEf − KEi)1, θ = 0°

Fd = ½mvf2

vf = 2Fd m

= 2(100N)(5.2m)50 kg

vf = 4.6 m/s

0, from rest!

A 50. kg sled is pulled for 5.2 m with a horizontal force of 100 N, starting from rest. Find the final speed. Ignore friction.

Work Changes EnergyAP

PHYS

Wnet, ext = ∆E; system: sled, m = 50kg

WFg + WFN + WF = ∆KE; A one body system can only have KE0 0 → because θ = 90°

0 + 0 + FdCosθ = (KEf − KEi)1, θ = 0°

Fd = ½mvf2

vf = 2Fd m

= 2(100N)(5.2m)50 kg

vf = 4.6 m/s

0, from rest!

AP

PHYS

A 1500 kg car is brought to a stop by a force of 7890 N in a distance of 20 m. How fast was the car moving?

burning transforms some of the chemical energy of hydrogen and oxygen into electromagnetic energy (light energy) and thermal energy.

Daily Quiz #5 — Work & Energy

A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little later it is moving at 8.2 m/s. How much did its height change between the two times? Assume friction and air resistance have no effect.

Wnet, ext = ∆E; system: car, m = 1200kg

WFg + WFN = ∆KE 0 → because θ = 90°

0 + FgdCos60° = (KEf − KEi)

Fgd(∆h ⁄ s) = ½mvf2 − ½mvi2

s

60°

Δh ⁄ s

∆h = ½mvf2 − ½mvi2

Fg

∆h = ½m(vf2 − vi2)mg

∆h = ½[ (8.2m/s)2 − (5m/s)2 ]9.8m/s2

∆h = 2.2 m

→ In the absence of non-conservative forces, mechanical energy is conserved.

Work Energy Practice problemA  1,200  kg  car  rolls  down  a  hill  with  its  brakes  off  and  transmission  in  neutral.  At  one  moment  it  is  moving  5.0  m/s.  A  little  later  it  is  moving  at  8.2  m/s.  How  much  did  its  height  change  between  the  two  times?  Assume  friction  and  air  resistance  have  no  effect.

AP

PHYS

Wnet, ext = ∆E; system: car, m = 1200kg & Earth

WFN = ∆E 0 → because θ = 90°

Fg is not an external force

0 = ∆E

A force is conservative if the work it does on a particle depends only on the initial and final positions of the particle.Conservative forces — Fg & Fsp = -kxNon-conservative force — Ffr

∆h = ½[ (8.2m/s)2 − (5m/s)2 ]9.8m/s2 → In the absence of

non-conservative forces, mechanical energy is conserved.

Work Energy Practice problemA  1,200  kg  car  rolls  down  a  hill  with  its  brakes  off  and  transmission  in  neutral.  At  one  moment  it  is  moving  5.0  m/s.  A  little  later  it  is  moving  at  8.2  m/s.  How  much  did  its  height  change  between  the  two  times?  Assume  friction  and  air  resistance  have  no  effect.

AP

PHYS

Wnet, ext = ∆E; system: car, m = 1200kg & Earth

WFN = ∆E 0 → because θ = 90°

0 = (KEf − KEi) − mg∆h

∆h = ½m(vf2 − vi2)mg

∆h = 2.2 m

Fg is not an external force

0 = ∆E

PE decreased

Suppose the spring and block are oriented vertically, as shown in the diagram. Initially, the spring is compressed 4.60 cm and the block is at rest. When the block is released, it accelerates upward. Find the speed of the block when the spring has returned to its equilibrium position.

Work Changes EnergyAP

PHYS

Wnet, ext = ∆E; system: block, Earth, Spring0 = ∆E

Fw

Fsp

0 = ∆KE + ∆PEg + ∆PEsp

0 = (KEf − KEi) + mg∆h − ½kx2 0, from rest!

k = 955 N/m0 = ½mvf2 + mg∆h − ½kx2

m = 1.70 kg

Suppose the spring and block are oriented vertically, as shown in the diagram. Initially, the spring is compressed 4.60 cm and the block is at rest. When the block is released, it accelerates upward. Find the speed of the block when the spring has returned to its equilibrium position.

Work Changes EnergyAP

PHYS

Fw

Fsp

k = 955 N/m

Wnet, ext = ∆E; system: block, Earth, Spring0 = ∆E

0 = ∆KE + ∆PEg + ∆PEsp

0 = (KEf − KEi) + mg∆h − ½kx2 0, from rest!

0 = ½mvf2 + mg∆h − ½kx2

½mvf2 = − mg∆h + ½kx2

vf2 = − mgd + ½kx2

½m

vf = − mgd + ½kx2

½kvf = − 1.70kg(9.8m/s2)(0.046m) + ½(955N/m)(0.046m)2

½(1.70 kg)

vf = 0.536 m/s