PHYS 218sec. 517-520

ReviewChap. 6

Work and Kinetic Energy

Work

sr

Displacement vector

Fr

Constant forceStraight-line displacement

Force and displacement are in the same direction.

, : magnitude of , : magnitude of W Fs F s= F sr r

Unit of work

1 joule (1 Newton) (1 meter) or 1J 1N m= ´ = ×

Work

Constant force in direction of straight-line displacement

W Fs=

Constant force, straight-line displacement

cosW Fs f= × =F sr r : force vector, : displacement vector,

: angle between the two vectorsf

F sr r

Varying x-component of force, straight-line displacement

2

1

x

xx

W F dx=ò

Work done on a curved path (general definition of work)

2 2 2

1 1 1

cosP P P

P P PW F dl Fdl df= = = ×ò ò ò FP

rrl

Evaluation of Work

Ex 6.1

ˆ ˆ ˆ ˆConstant force (160 N) (40 N) , displacement (14m) (11m) .

Then (160 N) (14m) ( 40 N) (11m) 1800 J

F i j s i j

W F s

= - = +

= × = ´ + - ´ =

r r

r r

Ex 6.2

( )( )

displacement 20 m, total weight 14,700 N,

constant force exerted by the tractor 5000 N, (angle 36.9 )

friction force 3500 N

TF

f

f

= =

= =

=

o

TF

n

w

f f ( )

work done by the tractor

cos (5000 N)(20 m)cos36.9 80,000J 80 kJ

work done by the friction force

cos180 (3500 N)(0 m) 1 70 kJ

work done by the normal force work done by the weight 0

angle

T T

f

W F s

W fs

f= = = =

= = - =-

= =

=

o

o

Q( )90

total work 10 kJT f n wW W W W W= + + + =

o

rs

Work done by a varying force, straight-line motion

m m

1x 2x

2xF

sr

1xF

1x 2x

( )xF x

1xF

2xF

1x 2x

1xF

2xF

axD

axF

bxD

bxF

2

1

To calculate the work,

divide the displacement into small segments , etc

is nearly constant for this small segment, and thena

x

ax a bx b xx

x

F

W F x F x F dx

D

= D + D + =òL

Work is the area under the curve between the initial and final positions in the graph of force as a function of position.

Evaluation of Work: Stretched spring

xxF kx=

xF kx=-

Hooke's law: , : spring constantF kx k=-

Restoring force

2 2

0

Work done by the force which stretches the spring ( )

when the elongation goes from 0 to a maximum value is

1 1 (work done by the spring is )

2 2

Generally, from

X

kx

X

W kx dx kX kX

=

= = -ò

1 2

2 22 1

to ,

1 1

2 2

x x x x

W kx kx

= =

= -

Varying x-component of force, straight-line displacement

2

1

x

xx

W F dx=ò

Evaluation of Work

( )2 2

1 12 1

x x

x x x xx x

W F dx F dx F x x F s= = = - =ò ò

When Fx is constant

2

1

In general, ,

so the work is obtained by calculating the above line integral,

which is defined on a given path.

P

PW d= ×ò F

rrl

Kinetic energy and work-energy theorem

Kinetic energy (K)

21

2K mv=

Work-Energy Theorem

When v = 0, K = 0

The work done by the net force on a particle equals the change in the particle’s kinetic energy

2 1totW K K K= - =DThis is a very general theorem. This theorem is true

regardless of the nature of the force.

Kinetic energy at the initial point

Kinetic energy at the final point

Proof of the work-energy theorem

Constant force in direction of straight-line displacement

m m

1x 2x

Fr

sr

1v 2v

2 22 2 2 12 1

2 22 22 12 1 2 1

For a constant force (acceleration ),

2 2

1 1

2 2 2

a

v vv v as a

s

v vW Fs mas m s mv mv K K

s

-= + Þ =

-Þ = = = = - = -

When a particle undergoes a displacement, it speeds up if W > 0, slows down if W < 0, and maintains the same speed if W = 0.

Proof of the work-energy theorem

Varying force, straight-line motion

2 2 2 2

1 1 1 1

2 22 1

Start from the definition of acceleration,

Then,

1 1

2 2 The work-energy theorem is valid.

x x x xx x x

x x x vx

tot x x x x xx x x v

dv dv dv dvdxa v v

dt dx dt dx dx

dvW F dx ma dx mv dx mv dv

dx

mv mv

= = = =

= = = =

= -

\

ò ò ò ò

Eliminate time dependence since W is an integral over x.

1 1 2 2At , and at ,x x v v x x v v= = = =

Ex 6.3 Same as Ex 6.2: suppose that the initial speed v1 is 2 m/s. What is the final speed?

Examples

2 1

We already know that 10kJ. To know the final speed, we use the work-energy theorem.

To compute kinetic energy, we need to know the mass. Since the total weight is 14,700 N,

14,700 N

tot

tot

W

W K K

wm

g

=

= -

= =

( )( )

2

221 1

2 1

2

22

1500 kg9.8 m/s

1 1Initial kinetic energy 1500 kg 2 m/s 3000 J

2 2Then ,

3000 J 10,000 J 13,000 J

Therefore,

2 2 13,000 J4.2 m/s

1500 kg

tot

K mv

K K W

K

Kv

m

=

= = =

= +

= + =

´= = =

Use Work-Energy Theorem to calculate speed.

Forces on a hammerheadEx 6.4

hammerhead

I-beam Point 3

Point 2

Point 1

3 m7.4 cm

Falling hammerhead

y

v

60 Nf =

w mg=

1 212 12

1 2 22 1 2 2 1

2

Total work done on the hammerhead: ( ) ( ) 5700J.

1Work-Energy Theorem gives ( 0)

2Therefore,

2 2 (5700 J)7.55 m/s

200 kg

tot

tot

tot

W w f s mg f s

W K K K mv v

Wv

m

®

®

= - = - =

= - = = =

´= = =

Q

Speed of the hammerhead at point 2

Forces on a hammerheadEx 6.4 (cont’d)

Hammerhead pushing I-beami.e., Point 2 Point 1y

60 Nf =

w mg=

n( )

( ) ( )

2 3 1 223 3 2

1 2

23

Total work done on the hammerhead:

5700 J1960 N 60 N 79,000 N

0.074 m

tot tot

tot

W w f n s K K W

Wn w f

s

® ®

®

= - - = - =-

Þ = - + = - + =

Work-energy theorem

Using the previous result

Be careful with the change of unit

Motion with a varying forceEx 6.7

m

1v

Friction (k)

k 10.1 kg, 20 N/m, 1.5 m/s,

Find the maximum distance if 0 and if 0.47,

when the object is moving from 0 to the rightk k

m k v

d

x

m m

= = =

= =

=

When 0km=

y

x

n

mg

springF

frF

2

22 1 1

2 21 1

1Work done by the spring

21

Then by Work-Energy theorem; 2

1 1Therefore, 10.6 cm

2 2

W kd

W K K mv

mkd mv d v

k

=-

= - =-

- =- Þ = =

When 0km ¹

2 21

Work done by the friction force

Then by Work-Energy theorem;

1 10.086 m

2 2

fr k

k

W mgd

mgd kd mv d

m

m

=-

- - =- Þ =

Ex 6.8 Motion on a curved path I

RRq

s Rq=

qdrl

y

x

Fr

q

T

F

mg

cosT q

sinT q

[ ]

0 0

0

0 0

0

0

The net force is zero ( remains in equilibrium)

sin 0, cos 0.

This leads to

tan

cos cos , and

Therefore,

tan cos ( ) sin

cos

(1 cos

x yF F T F T mg

F mg

F d Fdl F ds ds Rd

W mg Rd mgR d

mgR

mgR

q q

q

q q

q

q q q

q q q q q

q

q

= - = = - =

=

× = = =

= =

= -

= -

å å

ò ò

Q

rrl

)

Ex 6.9 Motion on a curved path II (another way to compute the line integral)

In Ex 6.8, can be written as

ˆ ˆcos sin .

All forces acting on the object are

ˆ ˆ ˆ ˆ( sin ) cos , ( ) , .

Therefore,

( sin )( cos ) ( cos )( sin ) 0

sin

d

d ds i ds j

T T i T j w w j F F i

T d T ds T ds

w d w ds

F d

q q

q q

q q q q

q

= +

= - + = - =

× = - + =

× =-

×

r

r

r rr

rr

rr

rr

l

l

l

l

l

0

cos tan cos sin

sin (1 cos )w

F w

F ds w ds w ds

W w d w ds wR

W W

q q q q

q q

= = =

Þ

= × = - =- -

=-

ò òrrl

We get the same answer as in Ex 6.8

Power

Average power

The time rate at which work is done.

av

WP

t

D=

D

(Instantaneous) power0

limt

W dWP

t dtD ®

D= =

D

Unit of powerwatt (W): 1W 1 J/s

horsepower (hp): 1 hp 746 W

=

=

Writing power in terms of F P F v= ×

r r