chapter 07: kinetic energy and work -...

Chapter 07: Kinetic Energy and Work

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.

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.

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....

Friction will convert mechanical energy to heat. Basically, this (conversion of mechanical energy to heat energy) is a non-reversible process.

Mechanical Energy is the sum of Kinetic energy + Potential energy. (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.m2/s2

Other useful unit of energy is the electron volt (eV)

1 eV = 1.602 x 10-19J

Work

Formal definition:

*Special* case: Work done by a constant force:

W = ( F cos θ) d = F d cos θ

Work done on an object moving with constant velocity? constant velocity => acceleration = 0 => force = 0

=> work = 0

Work is energy transferred to or from an object by means of a force acting on the object.

Component of F in direction of d

So, kinetic energy is mathematically connected to work!!

Consider 1-D motion.

v

(Integral over displacement becomes integral over velocity)

Work-Kinetic Energy Theorem

.... or in other words,

Final kinetic energy = Initial kinetic energy + net Work

The change in the kinetic energy of a particle is equal the net work done on the particle.

Wnet is work done by all forces

Work done by a constant force:

W = F d cos Ø = F . d

Consequences: (ø if the angle between F and d )

when Ø < 90o , W is positive

when Ø > 90o , W is negative

when Ø= 90o , 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

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

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.

d

Force – Displacement

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

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 Ø = F . d

•  W = F d cos ø = F . d

•  SI Unit for Work 1 N . m = 1 kg . m/s2 . m = 1 kg . m2/s2 = 1 J

•  Net work done by several forces ∆ W = Fnet

. d = ( F1 + F2 + F3 ) . d = F1

. d + F2 . d + F3

. d = W1 + W2 + W3

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?

a

v

How much work was done and when?

ds

Question: What about circular motion?

a

v How much work was done and when?

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 90o = 0, dot product is zero)

F

|vf | < |vi| means Kf < Ki, 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.

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?

x 0

v

Work done by Gravitation Force

Wg = mg d cos ø throw a ball upwards with v0: –  during the rise,

Wg = mg d cos180o = -mgd negative work K & v decrease

–  during the fall, Wg = mgd cos0o = mgd positive work K & v increase

mg

d

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 Wg is done on the crate by the gravitational force Fg during the lift?

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.

Fg is constant in this problem, so we can use W = Fg .d = mgd cos Ø, where Ø = 90o + θ

(a) How much work Wg is done on the crate by the gravitational force Fg during the lift?

Fg is constant in this problem, so we can use W = Fg .d = mgd cos Ø, where Ø = 90o + θ

Notice the triangle that is set up:

θ

d h = d sinθ

d cosθ

W = Fg .d = mgd cos(90o + θ) = -mgd sinθ = -mgh = -368 J

(The magnitude is equal to the increase in potential energy.)

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).

Notice, its all about h

How much work Wt is done on the crate by the force T from the cable during the lift?

vf = vi = 0 means ∆KE = 0, and the net work must equal zero

(Work – KE Theorem).

Wg = -368 J + WN = 0 J + WT = 0

because N d

WT = +368 J

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

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

Work-Kinetic Energy Theorem

.... or in other words,

final kinetic energy = initial kinetic energy + net work

The change in the kinetic energy of a particle is equal the net work done on the particle.

Work done by a variable force

•  Three dimensional analysis

The spring force is given by (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

F = -kx

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 | xf | > | xi | (further away from x = 0); W < 0 If | xf | < | xi | (closer to x = 0); W > 0

•  If xi = 0, xf = x then Ws = - ½ k x2

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: vi = vf = 0 then Ki = Kf = 0 work-kinetic energy theorem: Wa + Ws = ∆K = 0 therefore: Wa = - Ws

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/x2, with a = 9.0 Nm2

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 = 0o.

a is a constant, so it can be pulled out of the integral.

Problem: 07-79 Applications

Newton’s Law of Gravitation

Electrostatic Force between particles (Coulomb’s Law)

with G = 6.673 x 10-11 Nm2/kg2

with k = 8.988 x 109 Nm2/C2

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

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?

1) 0 m 2) 0.05 m 3) 0.5 m 4) 10 m 5) 100 m

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?


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

Initial Displacement (in meters) (t = 0s)

A) Final Displacement at t = 2.0s B) Work done on the particle

A particle of mass 5.0 kg experiences a time-dependent force (in Newtons) of:

Initial Velocity (t = 0s)

Vector representation Problem


A) Final Displacement at t = 2.0s



We know F so then we know the acceleration a






We know a so then we know the velocity v






We know v so then we know the displacement d



B) Work done on the particle



Two ways to approach this question

The easy way or The brute force way






Work = change in kinetic energy => W = Kf - Ki

The easy way

Now we need v(t=2s)







The easy way







The easy way

W = Kf - Ki = 252.9 - 122.5 = 130.4 J






The brute force way so

Power

•  The time rate at which work is done by a force •  Average power

Pavg = 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

Problem Force

Initial Displacement Final Displacement

A) Work done on the particle

B) Average power

C) Angle between di and df

In ∆t = 4.0s

Problem Force


A) Work done on the particle

in ∆t = 4.0s

F is constant in this problem, so

Problem Force


B) Average power

In ∆t = 4.0s

Pave = W/∆t = 32.0J/4s = 8.0W

Problem Force Initial Displacement Final Displacement

C) Angle between di and df

in ∆t = 4.0s

di and df are two vectors, so take the dot product

chapter 07: kinetic energy and work -...

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