Exam 1 distribution

Problem 3

SWWBSB vvv ///rrr

+=

v vB/WvB/WvW/S

vB/S

36

vB/W

vW/SvB/S

WB

SW

v

v

/

/36sin =WB

SW

v

v

/

/tan =

== 36sintan xxy

Problem 13 aN

mg

maNmg =N 735scale theof reading ==N

Rv

a2

=

Rvg

Nag

Nm 2

=

=

Problem 20

( ) ( ) 31 ttFm

ta == ( )ka ta=m

=f

i

t

t

zizfz dttavv )(,,

m/s 44

2

0

42

0

3===

=

=

t

t

z

tdttv543 22

22

=+=

=+= zx vvv

Work and Kinetic Energy

General Physics IMechanics

Racquetball Striking a Wall

Kinetic Energy MechanicsPhysics 140

Mt. EtnaStewart Hall

What is Work? In physics, work is a measure of how effective a

force is in moving things Work depends on both the force and how far

something moved under the influence of that force Work is a scalar Work is a scalar For a constant force F and a straight-line

displacement s, work is their scalar product:

SI unit of work is Joule: 1 J = 1 Nm = 1 kgm2/s2

cosFssFW == rr

F

s

Work Depends on the component of the force parallel

to the displacement:

||||cos FssFFssFW ==== rr

F Fs||

If components of both force and displacement are known, can also compute work by components:

F

sF||

F

s

zFyFxFsFW zyx ++==rr

Work Depends on the component of the force parallel

to the displacement:

Force along displacement does positive work Force against displacement does negative work Force perpendicular to displacement does no

work

Uniform Circular Motion A rock of mass m is twirled on a string in a

horizontal plane. The work W done by the tension in the string on the rock is

A. W < 0A. W < 0B. W = 0C. W > 0D. More than one option above is possible

Uniform Circular Motion In uniform circular

motion a force always acts perpendicular to the motion

Fr

vr

vr

Fr

the motion It has no component

along the motion This force does no

work!

vr F

r

vr

Fr

F

0=W

A block slides along a level surface subject to the forces shown. Which force does the largest work on the block during the displacement x? The magnitude of the forces is expressed by the length of their arrows.

N

F1

F

A. F1B. F2

mg

N

x

F2B. F2C. mgD. NE. Depends on mass

Work depends only on the component of force along the motion!

Lowering a book Yuri lowers a book of mass m downward a

distance h at constant speed v. The work done by the force of gravity is

A. NegativeA. NegativeB. ZeroC. PositiveD. Impossible to tell

Lowering a book Yuri lowers a book of mass m downward a

distance h at constant speed v. The work done by the force of Yuris hand is

A. NegativeA. NegativeB. ZeroC. PositiveD. Impossible to tell

Lowering a book Yuri lowers a book of mass m downward a

distance h at constant speed v. The work done by the total force acting on the book is

A. Negative HA. NegativeB. ZeroC. PositiveD. Impossible to tell

Wtot = WG + WH = 0G

H

s

a = 0

Work in General Situations What if force is not constant and trajectory is not

a straight line? Split the path into short displacements s Find the force F acting on the body over each s Find work W over each s by computing the Find work W over each s by computing the

scalar product:

Sum the results to find the total work W of the variable force over the curved path:

cossFsFW == rr

== cossFsFWrr

Work as Integral If the force varies

over a general path, add up many little Fds pieces:

= sdFWrr

This integral means exactly the same as the sum:

= sdFW

== sdFsFsFWrrrr cos

++== dzFdyFdxFdsFW zyx cos

Problem 6.34

(c): A. 20 J B. 40 J C. 60 J D. 80 J E. 120 J

If displacement is along xaxis, only the x-component of force Fx is relevant

If Fx varies with x, sum over small displacements

L++= xFxFW

Position-Dependent Force in 1D

In the limit of each x going to zero, this sum becomes an integral

Geometrically, work is the area under the Fx(x) curve

L++= bbxaax xFxFW

=2

1

x

xx dxFW

Position-Dependent Force in 1D An object is pushed by a variable force F plotted

below as a function of position x. How much work has the force done on the object when it has moved from x = 0 to x = + 6 m?

A. 0 Force, F

2 NA. 0B. 2 JC. 4 JD. 6 JE. 12 J

x

2 N

-2 N

6 m

+ +

_

Hookes Law For a spring, the force arising as a result of

deformation is proportional to the deformation If a spring is stretched or compressed from its

equilibrium length x0 by x = x x0, then the force arising in the spring is given by the Hookes law:

xkF =

Minus sign means that the force the spring exerts is opposite to the length change

xkFx =x0

x

x

F x

Hookes Law Coefficient k is called stiffness or stiffness

coefficient or force constant or spring constant of the spring

If choose x0 = 0 and extend the spring from 0 to x, then the work done by the spring is

( ) 2kxxx

The work done by an applied external force is exactly opposite:

( )2

2

00

kxdxkxdxFWxx

x ===

2

2kxWkxFx ==