kinetic energy 動能 and work 功 (chap. 7) what is energy? it is a conserved quantity 守怛量...

Kinetic energy 動能 and work 功 (Chap. 7)

What is energy?

it is a conserved quantity 守怛量 although energy can change form, it can be neither created nor destroyed.

this is called the principle of the conservation of energy.

when consider mechanical work, one can define energy as the capacity to do work.

in this chapter we focus on the relation between work and kinetic energy.

Kinetic energy

the energy associated with motion.

proportional to the mass and square of the speed of the object.

the unit is Joule 焦耳 (J=kg m2/s2).

1 calorie 卡路里 = 4.185 J

2

2

1mvK

For example, a 3 kg duck flying with speed 2m/s has kinetic energy of 6J.

v0

Work

An expression for work W:

Using the idea of conservation of energy, the increase of kinetic energy of an object must be equal to the work done by a force acting onto the object.

if KKW (work-kinetic energy theorem)

Given the force F (assume to be constant), the final v is given by:

dm

Fv

advv

)cos

(2

2

20

20

2

20

2

2

1

2

1cos mvmvFd

fK iK

dF

FdW

cosTherefore the work done by the force F is:

Note that the unit is:

J

skgm

mskgmNm

22

2

/

)/(

the unit of energyNo work is done if F is perpendicular to d:

In both cases, there is no change in kinetic energy (v is constant).

for example,

dFW

When the force F and displacement d is opposite, their dot product is negative, and hence a negative W. What does it mean?

If W is negative, the source of force F gains energy (does negative work) instead of loses energy.

(a) work done by friction is negative.

(b) work done by friction on A is positive.

dFFW 40cos30cos 21

0 Ng WW

20

2

2

1

2

1mvmvW f

0

Work done by gravitational force

• i.e. by its own weight

• which is always pointing towards the ground

• if a ball is moving upward, the work done by the gravity is negative.

• it means the ball loses kinetic energy (so v is reduced), and the gravity gains energy.

Energy and force are two different concepts, in this example the energy used is not very large, but the applied force is.

cosdFdFW ggg

dFg cos

cosgF

1)

hFdF gg cos2)

TdWT however we don’t know T (it may not be a constant).

Instead the total work done must be equal to the change in kinetic energy:

dF

K

N 0

0NW

gT WW So,

They are all the same: mghWg

Work done by a spring force

• a spring offers a variable force

• many physical force has similar nature

• to many springs, it is a good approximation to consider the spring force F as proportional to x:

kxF Hooke’s law

The minus sign indicates that the spring always try to return to its relaxed position.

Since F is changing with x, we cannot use

dFW

which is only valid for constant F.

k is the spring constant or force constant

Work done for non-constant force:

• one can assume the spring force Fi is approximately constant for a short displacement

• then the work done by the spring throughout this short displacement is still

• so the total work done from position xi to xf is simply the total sum of all work done:

iii xFW

ix

i

iii

i xFWW

Take the limit that 0 ix

f

i

x

xxdFW

The work is simply the area under the function F(x).

In components form:

dzFdyFdxFrdF

kdzjdyidxrd

kFjFiFF

zyx

zyx

ˆˆˆ

ˆˆˆ

Then put this into the work done formula gives us: f

i

f

i

f

i

z

z z

y

y y

x

x x dzFdyFdxFW

For the case of spring force, F=-kx and therefore is one-dimensional:

)(2

1

2

1

22

2

fi

x

x

x

x

x

x x

xxkW

kx

kxdx

dxFW

f

i

f

i

f

i

Work by spring force

22

2

1)0(

2

1kddkW

(b) Work done by the spring is

the negative sign means energy is stored in the spring.

22

2

1))(0(

2

1kddkW

(c)

energy is also stored in the spring

Obviously, at the moment when mass m is stopped, all its kinetic energy is transferred to the spring:

k

mvd

kdmv

22

2

1

2

1

(b), (a), (c), (d)

Power

• sometimes we are more interested in the rate of energy transfer than the work done itself.

• this rate is called the power.

• the average power is given by:

where Δt is the time taken by a force to do an amount of work W.

while the instantaneous power is:

The unit of P is J/s, or watt (W).

If the applied force is a constant, then one can rewrite the formula of power as:

vFP

dt

sdF

dt

sFd

dt

dWP

For circular motion, the applied force F is always perpendicular to the velocity v. Therefore P=0.

8m12

(a) 900J (the speed doesn’t affect the work done)

W=900J

(b) the force T of the rope = 900J/8m NT 5.112

Power

W

TvP

5.112)1(5.112

(c)

W

TvP

225)2(5.112

Problems: 3, 18, 28, 70

Since W=1/2 k x2 and from the graph we have:

2.0

)3(2

19.0 2

k

k

(a) JW 9.0)45)(2.0(2

1 22

(b) 1.2))2(5)(2.0(2

1 22 W (c) 0))5(5(2.02

1 22 W

FT

mgT

sin

cos

T

mg

22tan

dL

dmgmgF

22 dL

(a)

(b) Wtot=0=Kf - Ki

(c) )( 22 dLLmgWg

(d) WT = 0 (e)0 totgTF WWWW

22 dLLmgWF