In physics, work is done by forces on an object over a distance. 1a. W = F d (Joule, J) Force is parallel to distance Force is at an angle to distance 1b. W = F⋅d⋅Cosθ 2. W net = ∆E ext 3. Kinetic Energy, E k = 1/2m⋅v 2 (Joule, J) 4. Potential Energy, E P = m⋅g⋅h (Joule, J) 5. Power, P = W∕∆t (Watts, W) Chapter 6 Work & Energy A P P H Y S 5. Spring, E P = ½kx 2 (Joule, J)
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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
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