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Lecture 14

Physics I Chapter 10

Energy

Energy dissipation

Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

PHYS.1410 Lecture 14 Danylov Department of and Applied Physics IN IN THIS CHAPTER, you will learn to use the concepts of and .

Today we are going to discuss:

Chapter 10:  Energy Principle with Losses: Section 10.7-8

Chapter 11:  Newton’s 2nd law (more general form): Section 11.1  Linear Momentum: Section 11.1  Impulse: Section 11.1

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics Gravitational

Consider a block down on a frictionless surface under the influence of y y2 W   mgdy mg(y  y )  (U U ) K1 G  2 1 2 1  y1 ds The done by gravity depends only on coordinates of the final and initial positions, K so gravitational is conservative y 2 1 mg energy U  mgy Reference point y2 ̂ Actually, in general it is U  U 0  mgy ̂ Reference level U  0 x WG  U  U  W  K K U  K U Conservation of 2 2 1 1 Total Mechanical Energy E  K U

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics Example Roller coaster

The roller-coaster car starts from rest at the top of the hill. The height of the hill is 40 m. Calculate a) the of the car at the bottom of the hill; b) at what height it will have half this speed.

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics ConcepTest Water Slide I Paul and Kathleen start from rest at the A) Paul same on frictionless water slides B) Kathleen with different shapes. At the bottom, C) both the same whose is greater?

Conservation of Energy (for any of them): E  E i f i Ki Ui  K f U f therefore: 1 2 mgh  2 mv v  2gh because they both start from the same height (h), they have the same velocity at the bottom. f Ref. level U=0 Spring Potential Energy

Fsp  kx What is the potential energy of a spring compressed from equilibrium by a distance x?

Work done by a spring k 2 2 (from the previous class, Wsp   x f  xi  Let’s combine Lecture 13) 2 them Use a relation between potential energy and work:

k  (U U )   x 2  x 2 f i 2 f i

1 2 Potential energy From here you can see that the PE of a spring is U  kx spring 2 of a spring Where x is a from an equilibrium of a spring

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics Example Brick/spring on a track

A 2 kg , with an initial velocity of 5 m/s, slides down the frictionless track shown below and into a spring with spring constant k=250 N/m. How far is the spring compressed?

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics How to solve problems using the energy approach if there are losses of mechanical energy in a system due to presence of a force?

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics with nonconservative

Consider an object experiences conservative and nonconservative forces:    Fnet  FC  FNC Since the work-KE principle works for ANY forces, the total work done on the object

Wnet  K  y FC W W  K C NC  f Recall, from the previous class, if a  ds force is conservative, then  U  WC FNC

 U WNC  K  (U U ) W  K  K i f i NC f i x

K f U f  Ki Ui WNC E f  Ei  WNC

Total Mech Energy Total Mech Energy Mech. Energy converted to at the final point at the initial point (losses) We are going to have a friction force   (Work done by friction as a nonconservative force W fr  fk  s is usually negative) PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics Example Block sliding down Inclined Plane

A block of mass m slides down a plane of length l and angle α. Find the speed of the block at the bottom of the incline plane assuming that it starts from rest and the coefficient of friction μ is constant.

 fk

(Thermal energy is generated)

α α mg

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics Review

1 2 K  2 mv

Gravitational U  mgy potential energy

1 2 Potential energy of a spring Uspring  kx 2 Total Mechanical Energy E  K U Conservation of Mechanical Energy

Without losses (no friction) K f U f  Ki Ui

With losses (with friction) K f U f  Ki Ui WNC

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics ConcepTest Water Slide II

Paul and Kathleen start from rest at A) Paul the same time on frictionless water B) Kathleen slides with different shapes. Who C) both the same makes it to the bottom first?

Even though they both have the same final velocity, Kathleen is at a lower height than Paul for most of her ride. Thus, she always has a larger velocity during her ride and therefore arrives earlier!

Ref. level U=0 http://phys23p.sl.psu.edu/phys_anim/mech/ramped.avi Thank you

PHYS.1410 Lecture 14 Danylov Department of Physics and Applied Physics