Chapter 07: Kinetic Energy and Work

Chapter 07: Kinetic Energy and Work Conservation of Energy is one of Nature s fundamental laws that is not violated. Energy can take on different forms in a given system. This chapter we will discuss work and kinetic energy. Next chapter we will discuss potential energy. If we put energy into the system by doing work, this additional energy has to go somewhere. That is, the kinetic energy increases or as in next chapter, the potential energy increases. The opposite is also true when we take energy out of a system.

Forms of energy Kinetic energy: the energy associated with motion. Potential energy: the energy associated with position or state. (Chapter 08) Heat: thermodynamic quanitity related to entropy. (Chapter 20) Conservation of Energy is one of Nature s fundamental laws that is not violated. That means the grand total of all forms of energy in a given system is (and was, and will be) a constant.

Different forms of energy Kinetic Energy: Potential Energy: linear motion gravitational rotational motion spring compression/tension electrostatic/magnetostatic chemical, nuclear, etc... Mechanical Energy is the sum of Kinetic energy + Potential energy. (reversible process) Friction will convert mechanical energy to heat. Basically, this (conversion of mechanical energy to heat energy) is a non-reversible process.

Chapter 07: Kinetic Energy and Work Kinetic Energy is the energy associated with the motion of an object. m: mass and v: speed Note: This form of K.E. holds only for speeds v << c. SI unit of energy: 1 joule = 1 J = 1 kg.m 2 /s 2 Other useful unit of energy is the electron volt (ev) 1 ev = 1.602 x 10-19 J

Formal definition: Work Work is energy transferred to or from an object by means of a force acting on the object. *Special* case: Work done by a constant force: W = ( F cos θ) d = F d cos θ Component of F in direction of d Work done on an object moving with constant velocity? constant velocity => acceleration = 0 => force = 0 => work = 0

Consider 1-D motion. v (Integral over displacement becomes integral over velocity) So, kinetic energy is mathematically connected to work!!

Work-Kinetic Energy Theorem The change in the kinetic energy of a particle is equal the net work done on the particle.... or in other words, Final kinetic energy = Initial kinetic energy + net Work W net is work done by all forces

Work done by a constant force: W = F d cos = F. d Consequences: (ø if the angle between F and d ) when < 90 o, W is positive when > 90 o, W is negative when Ø= 90 o, W = 0 when F or d is zero, W = 0 Work done on the object by the force: Positive work: object receives energy Negative work: object loses energy

Force displacement question In all of the four situations shown below, the box is sliding to the right a distance of d. The magnitudes of the forces (black arrows) are identical. Rank the work done during the displacement, from most positive to most negative. 1. a, b, c, d d 2. d, c, b, a 3. b, a, c, d 4. all result in the same work 5. none of the above

Force Displacement In all of the four situations shown below, the box is sliding to the right a distance of d. The magnitudes of the forces are identical. Rank the work done during the displacement, from most positive to most negative. W = F d cos = F. d Rank +W -W 1) a,b,c,d 2) d,c,b,a 3) b,a,c,d 4) all result in the same work 5) none of the above d

W = F d cos = F. d SI Unit for Work 1 N. m = 1 kg. m/s 2. m = 1 kg. m 2 /s 2 = 1 J Net work done by several forces W = F net. d = ( F 1 + F 2 + F 3 ). d = F 1. d + F 2. d + F 3. d = W 1 + W 2 + W 3 Remember that the above equations hold ONLY for constant forces. In general, you must integrate the force(s) over displacement!

Question: What about circular motion? How much work was done and when? v a ds

Question: What about circular motion? How much work was done and when? a v F Work was done to start the motion and then NO work is required to keep the object in circular motion! Centripetal force is perpendicular to the velocity and thus the displacement. (F and d vectors are perpendicular, cos 90 o = 0, dot product is zero)

A particle moves along the x-axis. Does the kinetic energy of the particle increase, decrease, or remain the same if the particle s velocity changes? (a) from 3 m/s to 2 m/s? (b) from 2 m/s to 2 m/s? v f < v i means K f < K i, so work is negative 2 m/s 0 m/s: work negative 0 m/s +2 m/s: work positive Together they add up to net zero work. v 0 x

Work done by Gravitation Force W g = mg d cos throw a ball upwards with v 0 : during the rise, W g = mg d cos180 o = mgd negative work K & v decrease mg d during the fall, W g = mgd cos0 o = mgd positive work K & v increase mg d Final kinetic energy = Initial kinetic energy + net Work

Sample problem 7-5: An initially stationary 15.0 kg crate is pulled a distance L = 5.70 m up a frictionless ramp, to a height H of 2.5 m, where it stops. (a) How much work W g is done on the crate by the gravitational force F g during the lift?

F g is constant in this problem, so we can use W = F g. d = mgd cos, where = 90 o + θ Notice the triangle that is set up: d h = d sinθ θ d cosθ Notice, its all about h W = F g. d = mgd cos(90 o + θ) = mgd sinθ = mgh = 368 J Seems funny to say this negative work is done by the force of gravity. Instead, we say this negative work was done against gravity (by some other force). (The magnitude is equal to the increase in potential energy.)

How much work W t is done on the crate by the force T from the cable during the lift? v f = v i = 0 means KE = 0, and the net work must equal zero (Work KE Theorem). W g = 368 J + W N = 0 J + W T = 0 W T = +368 J because N d

Sample Problem A batter hits a baseball to the outfield. Which graph could represent its kinetic energy as a function of time? 1 2 3 4 5 6 7 8 9) none of the above

A batter hits a baseball to the outfield. Which graph could represent its kinetic energy as a function of time? 1 2 3 4 5 6 7 8 9) none of the above

Work-Kinetic Energy Theorem The change in the kinetic energy of a particle is equal the net work done on the particle.... or in other words, final kinetic energy = initial kinetic energy + net work

Work done by a variable force Three dimensional analysis

Work done by a spring force The spring force is given by F = kx (Hooke s law) k: spring (or force) constant k relates to stiffness of spring; unit for k: N/m. Spring force is a variable force. x = 0 at the free end of the relaxed spring.

Work done by a spring force The spring force tries to restore the system to its equilibrium state (position). Examples: springs molecules pendulum Motion results in Simple Harmonic Motion (Chapter 15)

Work done by Spring Force Work done by the spring force If x f > x i (further away from x = 0); W < 0 If x f < x i (closer to x = 0); W > 0 If x i = 0, x f = x then W s = ½ k x 2 This work went into potential energy, since the speeds are zero before and after.

The work done by an applied force (i.e., us). If the block is stationary both before and after the displacement: v i = v f = 0 then K i = K f = 0 work-kinetic energy theorem: W a + W s = K = 0 therefore: W a = W s

Problem 07-79 A force that is directed along the positive x direction acts on a particle moving along the x- axis. The force is of the form: F = a/x 2, with a = 9.0 Nm 2 Find the work done on the particle by the force as the particle moves from x=1.0m to x = 3.0m.

Problem: 07-79 Solution F in +x direction, displacement also in +x direction => angle between F and x = 0 o. a is a constant, so it can be pulled out of the integral.

Problem: 07-79 Applications Newton s Law of Gravitation with G = 6.673 x 10 11 Nm 2 /kg 2 Electrostatic Force between particles (Coulomb s Law) with k = 8.988 x 10 9 Nm 2 /C 2

Spring Oscillation A 2.0 kg block is attached to a horizonal ideal spring with a spring constant of k = 200 N/m. When the spring is at its equilibrium position, the block is given a speed of 5.0 m/s. What is the maximum displacement (i.e., amplitude) of the spring? 1) 0 m 2) 0.05 m 3) 0.5 m 4) 10 m 5) 100 m

Spring Oscillation 1) 0 m 2) 0.05 m 3) 0.5 m 4) 10 m 5) 100 m A 2.0 kg block is attached to a horizonal ideal spring with a spring constant of k = 200 N/m. When the spring is at its equilibrium position, the block is given a speed of 5.0 m/s. What is the maximum displacement (i.e., amplitude, A) of the spring? 0.0 Work Kinetic Energy Theorem

Problem 07-54 A 0.25 kg block is dropped on a relaxed spring that has a spring constant of k = 250.0 N/m (2.5 N/cm). The block becomes attached to the spring and compresses it 0.12 m before momentarily stopping. While the spring is being compressed, what work is done on the block by: A) Gravity (i.e. the gravitational force on it)? B) The spring force? C) What is the speed of the block just prior to hitting the spring?

Vector representation of a Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) A) Final Displacement at t = 2.0s B) Work done on the particle

Vector representation Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) A) Final Displacement at t = 2.0s We know F so then we know the acceleration a

Vector representation Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) B) Work done on the particle Two ways to approach this question The easy way or The brute force way

Vector representation Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) B) Work done on the particle The easy way Work = change in kinetic energy => W = K f K i Now we need v(t=2s)

Vector representation Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) B) Work done on the particle The easy way Work = change in kinetic energy => W = K f K i

Vector representation Problem A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of: Initial Displacement (in meters) (t = 0s) Initial Velocity (t = 0s) B) Work done on the particle The easy way Work = change in kinetic energy => W = K f K i W = K f K i = 252.9 122.5 = 130.4 J

Power The time rate at which work is done by a force Average power P avg = W/ t (energy per time) Instantaneous power P = dw/dt = (Fcosθ dx)/dt = F v cosθ= F. v Unit: watt 1 watt = 1 W = 1 J / s 1 horsepower = 1 hp = 550 ft lb/s = 746 W kilowatt-hour is a unit for energy or work: 1 kw h = 3.6 M J

Force Initial Displacement Problem In t = 4.0s Final Displacement A) Work done on the particle B) Average power C) Angle between d i and d f

Force Initial Displacement Problem in t = 4.0s Final Displacement A) Work done on the particle F is constant in this problem, so

Force Initial Displacement Problem In t = 4.0s Final Displacement B) Average power P ave = W/ t = 32.0J/4s = 8.0W

Force Initial Displacement Problem in t = 4.0s Final Displacement C) Angle between d i and d f d i and d f are two vectors, so take the dot product