Kinetic Energy, Work, and Power

PHY11 Lesson 1

Definition of Energy

• It is a (scalar) quantity that can be converted from one form to another but cannot be created nor destroyed.

• It is a quantity that can transferred from one body to another.

• It is the capacity of a body to do work.

Units of Energy

1 joule (J) = 1 N.m 1 ft-lb = 1.356 J

1 J = 107 ergs

1 J = 0.239 cal

1 Btu = 1055 J

1 eV = 1.602 x 10-19 J

Kinetic Energy (K)

• It is the energy associated with the state of motion of an object.

Conservation of Energy

• The total energy of a (closed) system does not change.

Consider an object of mass m moving with a speed v. The kinetic energy of the object is

21

2K mv

Work (W)

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

Consider a constant force F acting on an object of mass m that is free to move across a frictionless horizontal surface.

m

F

If the force causes the object to move through a displacement d, then the work done by the force on the object is

m

F

d

cosW Fd W F d ############################

Work and Kinetic Energy

Consider that, at t = 0, a constant force F acts on an object of mass m and initial velocity vo.

m F

vo

m F

d

v

From Newton’s 2nd Law: F ma Recall:

2 2

2ov v

ad

At time t > 0, the object would have a displacement d and a final velocity v.

2 2

2ov v

F md

2 21 1

2 2 oFd mv mv

Work-Kinetic Energy Theorem

2 21 1

2 2 oFd mv mv

work done

finalkineticenergy

initialkineticenergy

The total work done by a force (or forces) on an object is equal to the change in the kinetic energy of the object.

KKKW o

Problem 1

A bead with a mass 0.012 kg is moving along a wire in the positive direction of an axis. Beginning at time t=0, when the bead passes through x=0 with a speed 12 m/s, a constant force acts on the bead. The figure below indicates the bead’s position at these four times: to = 0, t1 = 1.0 s, t2 = 2.0 s, and t3 = 3.0 s. What is the kinetic energy of the bead at t = 10.0 s?

x(m)

0 5 10 15 20

to t1 t2 t3

Problem 1: Solution

The motion of the bead is UAM and its displacement is given by

From the figure, x = 10m when t1 = 1.0 s

22112)( atttx

221 )1()1(1210 a 2/0.4 sma

2212)( tttx tdt

dxv 412

At t = 10.0 s, smv /28)10(412

At t = 10.0 s, JsmkgK 7.4)/28)(012.0( 221

Problem 2

A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

Answers: (a) 2.4 m/s (b) 4.8 m/s

Problem 3

A 4.2 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x-axis along the track is applied to the body. The positions of the body as it slides to the right is shown in the figure. The force F is applied to the body at t = 0, and the graph records the position of the body at t = 1.00 s and t = 2.00 s. How much work is done on the body by the applied force  between t = 0 and t = 1.5 s?

Answer: 0.756 J

x(m)

0 0.2 0.4

0.6

0.8

t=0

1.0s 2.0s

Work Done by Gravitational Force Fg

Consider an object of mass m thrown upward with an initial velocity vo and initial kinetic energy Ko =½ mvo

2.

vo

Fg

v

Fgd

As the object rises to a height d the work done by the gravitational force Fg is

180cosmgdW

mgdW

As an object rises, gravitational force does negative work on it and its kinetic energy decreases.

Work Done by Gravitational Force Fg

When the object has reached its maximum height, it will fall back down.

vo=0

Fgd

0cosmgdW

mgdW

As an object falls down, gravitational force does positive work on it and its kinetic energy increases.

Fg

v

The work done by gravitational force becomes

Work Done in Lifting/Lowering an Object

F

Fg

d

FdWa

As the object rises, the work done by the applied force is

Consider an object of mass m lifted by a force F.

F

Fg

The work done gravity is

mgdWg From the work-kinetic energy theorem,

gao WWKKK

Work Done in Lifting/Lowering an Object

F

Fg d

FdWa

If the object is being lowered, the work done by the force is

F

Fg

The work done gravity is

mgdWg

As an object is being raised, the work done by the applied force is positive while that of gravitational force is negative.

As an object is being lowered, the work done by the applied force is negative while that of gravitational force is positive.

Work Done by a Spring Force

Consider an ideal spring in its relaxed state (x=0).

where k is the called the spring constant or (force constant).

• The spring force is not constant.

• If Fx is (+) then x is (-) and vice versa.

kxFx

If a force is applied to stretch (or compress) the spring, the spring force is given by Hooke’s Law.

Work Done by a Spring Force

Consider an ideal spring (massless and obeys Hooke’s Law) of force constant k and a spring force Fx.

xFW aves

The spring force is a variable force and is directly proportional to x and the work done is

221 kxWs

2212

21 kxkxW os

xF

W xs

2

0

2221

os xxkW

Work Done by an Applied Force Consider a block of mass m attached to an ideal spring of force constant k.

m

Ki

m

x

Kf

FaFs

KKKWW osa

If there is no change in the kinetic energy of the block,

sa WW

Suppose a force Fa displaces the block along the x-axis. The work done by this applied force Wa plus the work done by the spring force Ws give

Work Done by a General Variable Force

Consider a variable force acting on an object in the +x-direction and that the force varies with the position x.

The work done by such variable force is

x

xodxxFW )(

Work Done by a Spring Force

)(2

1 22o

x

x

x

x xs xxkdxxkdxFWo

f

i

2212

21 kxkxW os

Work-Kinetic Energy Theorem with a Variable Force

Consider a variable force acting on an object in the +x-direction and that the force varies with the position x.

The work done by such variable force is

x

x

x

x

x

x ooo

dxdt

dvmdxmadxxFW )(

vdx

dv

dt

dx

dx

dv

dt

dv

v

v

v

v oo

dvvmdvmvW

KKKW o 22

2

1

2

1omvmvW

Work: Area Under the Curve

)( oxxFW

Constant Force

F(x)

xxo x

F

W

2

2

1

2

1

2

1kxxFbhW

Variable Force (F x)

F(x)

xx

W

F

General Variable Force

x

xodxxFW )(

xo x

Answers: (a) Fworker =74 N(b) Wworker =+333 J(c) Wf = -333 J(d) WN =0, Wg =0(e) WNet =0

Problem 4

Answers: (a) 30.3 m/s(b) 46.8 m

Problem 5

Answer: 2.90 m/s

Problem 6

the figure.

(see figure.)

Problem 7

Answers: (a) 0.600 m(b) 1.50 m/s

Problem 8

Answer: -209 J

Power

Power is the time rate at which work is done by a force.

t

WPavg

Average Power is the work done W by a force F in an amount of time t.

Instantaneous Power is the instantaneous time rate of doing work.

dt

dWP

Units of Power: slbftsJWwatt /.738.0/111

slbftWhphorsepower /.55074611

Power and Velocity

dt

dxF

dt

dWP

cos

cosFvP

vFP

Problem 9

Answer: 390 W

Problem 10

Answer: 28 passengers