2nd Midterm

Friday, February 20, 2014 4-5:50 p.m.

amF

Newton’s second law, vector form

yyxx maF ,maF Newton’s second law,

component form

Chapter 5 Summary

nf and nf KKsmax,s

Two types: Static Friction and Kinetic Friction

r

vac

2

r

vmmaF c

2

(5.6) Applied to a Particle in Uniform Circular Motion

(5.5)

Although it came originally from Chapter 4, you need to know

Newton’s Second Law:

Work is done only when:

1. a force is applied to a body

2. the body moves

Examples of work not being done are

1. Pushing against a building and hanging from a cliff do not

constitute work, as no displacement is involved.

2. Holding a heavy box at arms’ length is not work, as there is no

displacement.

6. Work and Kinetic Energy

Summary of Work Formulas

f

i

f

i

r

r

x

xx

rdFW

dxFW

xFW

rFW

rFW

Constant force

Non-Constant force

with several constant forces

with several 1D constant forces

1D non-constant force(s)

General Case

f

i

net

r

r

dW rF

The force of a spring varies with distance and is given by

Hooke's law:

kxF

22

22

2

1

2)(

fi

if

x

x

x

x

ss

xxkW

xxk

dxkxdxFW

f

i

f

i

The kinetic energy of a particle of mass m moving with a

speed v is

2

21mvK

The work-kinetic energy theorem states that when work

is done on a system and the only change in the system is in

its speed, the net work done on the system by external

forces equals the change in kinetic energy of the system:

22

21

21

ififmvmvKKK

netW

In the case of an object sliding through displacement d over

a surface with friction, the work done by friction is:

dfW kf

The change in internal energy (heating) is given by

negative value of the work done by friction:

dfWE kf int

Average power is the time rate of energy transfer. If we

use work as the energy transfer mechanism,

If an agent applies a force F to an object moving with a

velocity v, the instantaneous power delivered by that

agent is

tWP

vF dt

dWP

Solution 6.84

Solution 6.91

Example 1

A skier of mass 70.9 kg is pulled up a slope by a motor-driven cable.

(a) How much work is required to pull him a distance of 60.0 m up a

29.0° slope (assumed frictionless) at a constant speed of 2.07 m/s?

(b) A motor of what power is required to perform this task?

Example 1 solution

0

K

KW

The skier moves at constant speed.

kJmsmkgW

mgdWW

o

M

gM

2.200.29sin)0.60)(/80.9()9.70(

)sin(

2

hpWs

P

t

WP

ssm

m

speed

dt

M

934.069729

102.20

29/07.2

0.60

3

(b) A motor of what power is required to perform this task?

Chapter 7 Summary

A force is conservative if the work it does on a particle is

independent of the path the particle takes between two

given points. A conservative force in mechanics does not

cause a transformation of mechanical energy to internal

energy. A force that does not meet these criteria is said to

be non-conservative.

If a particle of mass m is elevated a distance y from a

reference point y = 0 near the Earth’s surface, the

gravitational potential energy of the particle-Earth system

can be defined as

mgygU

The elastic potential energy stored in a spring of force

constant k that is compressed or stretched a distance x is

2

2

1kxsU

A potential energy function U can be associated only with a

conservative force. If a conservative force F acts on a

particle that moves along the x axis from xi to xf, the

change in the potential energy equals the negative of the

work done by that force:

fx

ixdxxFiUfU

The total mechanical energy of a system is defined as the sum of

the kinetic energy and potential energy:

UKE mech

The principle of conservation of mechanical energy of the system

is constant if all the forces in the system are conservative:

0 UK

ffiiUKUK

fiEE

If some of the forces acting with a system are not

conservative, the mechanical energy of the system does not

remain constant and the energy equation becomes:

int0K U E

In the case of friction being the non-conservative force, the

last term is due to the work done by friction, which is the

negative of the force of friction (fk) times the distance

traveled (d), i.e.,

nc k

K U W f d

Problem

A 20.0 kg block is connected to a 30.0 kg block by a string that

passes over a light, frictionless pulley. The 30.0 kg block is

connected to a spring that has negligible mass and a force

constant of 280 N/m, as shown in the figure below. The spring is

unstretched when the system is as shown in the figure, and the

incline is frictionless. The 20.0 kg block is pulled 16.0 cm down

the incline (so that the 30.0 kg block is 36.0 cm above the floor)

and released from rest. Find the speed of each block when the

30.0 kg block is 20.0 cm above the floor (that is, when the spring

is unstretched).

fii UKfUK

sin111 gdmhhgmU BAgr For block 2 the change in the potential energy is:

2222

2

1 and kdUgdmU sgr

smv

msmkg

v

kdmmgdmm

v

kdmmgdvmm

kdgdmgdmvmvm

/1.1

)16.0(2802

140sin2030)16.0)(/8.9(

)3020(

2

2

1sin

2

2

1sin

2

02

1sin

2

1

2

1

22

2

12

21

2

12

221

2

21

2

2

2

1

A

B

d

For block 1 the change in the potential energy is:

Rotation of Rigid Objects: Rotational Kinematics

The Rigid Object under Constant Angular Acceleration

f i t

12

f i f i t

f i i t 12 t 2

f

2 i

2 2 f i

at is the tangential component of the linear acceleration of the

particle moving in a circle of radius r

is the angular acceleration of the rotating body

Centripetal, acceleration to be ar = v2/r for a particle

moving in a circle.

v is the linear speed of the particle moving in a circle of radius r

is the angular speed of the rotating body

We define I the rotational inertia or moment of inertia as

The unit of the moment of inertia is kg.m2.

Kinetic energy of rotation

whereris the density of the object and V is its volume.

Moment of inertia for a continuous rigid object

For the one-dimensional case dm = ldx where l is mass per unit length

dxdm /l dxxI 2l

For the two-dimensional case dm = σds where σ is mass per unit area

dsdm / dsrI 2

Parallel axis theorem

Moments of inertia of rigid objects of uniform density with different

shapes

Example: A spring with spring constant k is used to propel an object of

mass m along a horizontal surface. The spring is compressed a distance

d before being released. Beneath the spring the surface is frictionless,

but at some point to the right of the spring, the surface becomes rough

with coefficient of kinetic friction k. After being released, the mass

moves away from the spring and encounters the rough surface. Express

your answers in terms of k, m, d, mk and g.

a) What is the speed of the mass after it leaves the spring, but before it

hits the rough patch?

b) What is the total work done by the spring on the mass?

c) How far along the rough surface will the object travel before coming

to rest?

d) What is the total work done by friction after the object comes to rest?

k

m k

d

22

2

1

2

1mvkd

a)

m

kdv

b) WkdkxkxdxW

dd

20

02

2

1

2

1

c) fWE

mgxxfW

E

kdE

kkf

f

i

0

2

1 2

mg

kdx

mgxkd

k

k

2

2

10

2

2

use in the energy equation

22

2

1

2kd

mg

kdmgmgxW

kkkf

d)