PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power

PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power

Exam #1 • Thursday, September 29, 7:00 – 8:00 PM. • Material: Chapters 2-6.3 • Use black lead #2 pencil and calculator. • Formula sheet provided. You may bring 1 extra sheet of handwritten notes (both sides) but no other materials • Sample exams on Web (link on homepage) Room assignments - to be announced.

Supplemental Instruction Exam Review Session Tuesday Sept. 27th 7-9pm RHPH 172

Reading Quiz: Sections 6.2 & 6.3 During a short time interval a particle moves along a straight line a distance • The work done on the particle was • 0 • 4 • 8 • 16 • . During that time a constant force acted on the particle:

Two marbles, one twice as heavy as the the other, are dropped to the ground from the roof a building. Just before hitting the ground, the heavier marble has (a) as much kinetic energy as the lighter one (b) twice as much kinetic energy as the lighter one (c) half as much kinetic energy as the lighter one (d) four times as much kinetic energy as the lighter one (e) impossible to determine

T v v N Review: Constant Force... W = Fr • If = 90o no work is done. No work done by T No work done by N

Work-Kinetic Energy Theorem Work done by the net external (constant) force equals the change in kinetic energy {NetWork done on object}= {change in kinetic energy of object}

shelf FHAND r v = const a= 0 mg floor Example: Lifting a book from the floor to a shelf Work done by Lifting • First calculate the work done by gravity: Wg = mgr= -mg r • Now find the work done bythe hand: WHAND = FHANDr= FHANDr

shelf FHAND r v = const a= 0 mg floor Work done by Lifting Work/Kinetic Energy Theorem: W = K When lifting a book from the floor to a shelf, the object is stationary before and after the lift: WNET= 0 WNET = WHAND + Wg = FHAND r- mg r = (FHAND- mg)r WHAND = - Wg

shelf FHAND r v = const a= 0 shelf FHAND r v = const a= 0 mg floor mg floor Lifting vs. Lowering Lifting Lowering Wg= -mg r Wg= mg r WHAND= FHANDr WHAND= -FHANDr WHAND = - Wg WHAND = - Wg

= Fr1+ Fr2+ . . . +Frn = F(r1+ r2+ . . .+ rn) r1 =Fr = Fy r2 r r3 Wg = mgh rn Work done by gravity... W NET = W1 + W2 +. . .+ Wn m mg h j

F(x) x1 x2 dx Work done by Variable Force: (1D) F * When the force was constant, we wrote W = Fx area under F vs. x plot: Wg x x * For variable force, we find the area by integrating: dW = F(x) dx.

dx dv dv dx v dx v dv v22 v12 v22 v12 Work/Kinetic Energy Theorem for a Variable Force F dv F dx dv dv dv v (chain rule) dx = = dx dt dx dt

Power Power is the rate at which work is done by a force PAVG = W/Dt Average Power P = dW/dt Instantaneous Power The unit of power is a Joule/second (J/s) which we define as a Watt (W) 1 W = 1 J/s

The force on a particle of mass m is given by Choose the correct statement: The work done will be the same going from x=1 to x=2 as it is for going from x=0 to x=1. The work done will be the same going from x=1 to x=2 as it is for going from x=-1 to x=-2. The average power is the same as the instantaneous power. None of the above are correct.

Spring unstretched Person pulling Person pushing Fs Fp Fs Fp Work done by a Spring x=0 *For a person to hold a spring stretched out or compressed by x from its unstretched length, it requires a force where k =spring constant measures the stiffness of the spring.

Spring unstretched Person pulling Person pushing Fs Fp Fs Fp Hooke’s law Work done by a Spring x=0 The spring exerts a force (restoring force) in the opposite direction: where k =spring constant measures the stiffness of the spring.

1-D Variable Force Example: Spring *For a spring Fx = -kx. ( Hooke’s Law) k =spring constant F(x) x1 x2 x relaxed position -kx F= - k x1 F= - k x2

1-D Variable Force Example: Spring *The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws relaxed position -kx F= - k x1 F= - k x2

1-D Variable Force Example: Spring *The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws -kx

x v1 m1 m1 Work - Energy A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest.

x1 v1 m1 m1 Work - Energy Use the fact that WNET = DK. In this case WNET =WSPRING = -1/2 kx2and K = -1/2 mv2 so kx2 = mv2 In the case of x1

(a) (b) (c) x2 v2 m2 m2 Work - Energy If the initialspeed of the box were doubled and its mass were halved, how far x2 would the spring compress ?

x2 v2 m2 m2 Work - Energy If the initialspeed of the box were doubled and its mass were halved, how far x2 would the spring compress ? So if v2 = 2v1 and m2 = m1/2

A person pulls on a spring. It requires a force of 75N to stretch it by 3 cm. How much work does the person do? If the person compresses the spring by 3 cm how much work does the person do? Calculate the spring constant: The work is The work to compress the spring is the same since W is proportional to x2. Example

A horizontal spring has k=360N/m. (a) How much work is required to compress a spring from x=0 to x=-11 cm? (b) If a 1.85 kg block is placed against the spring and the spring is released what will be the speed of the block when it separates from the spring at x=0? Example: Compressed Spring N Fs=kx x=0 x=-11 mg

Example: Compressed Spring The work done to stretch or compress the spring is: In returning to its uncompressed length the spring will do work W=2.18J on the block. According to the work-energy principle the block acquires kinetic energy:

PHYS 152 Fall 2005 Thursday, Sep 22 Chapter 6 - Work, Kinetic Energy & Power