solution+test+fyanvc09++ch+3,+4,+8 4,+9 2 5,+10!1!6+vectors,+newton's+laws,+torque,+pressure

8/3/2019 Solution+Test+FyANVC09++Ch+3,+4,+8 4,+9 2 5,+10!1!6+Vectors,+Newton's+Laws,+Torque,+Pressure

1/19


2/19

Solution FyANVC09 Ch 3, 4, 8.4, 9.2-5, 10.1-6 Vectors, Newton's laws, Torque, Pressure NV-College

[email protected] to use for educational purposes.Not for sale! page 2 of 19

In the multiple choice problems below, circle the correct alternative and write clearly the

correct answer in the space provided as Alternative:

1. The diagram below shows a N510 truck at rest on a hill that makes an angle of

8.0 with the horizontal. What is the component of the trucks weight parallel to the hill?

0.1

a) kN99

b) kN14

c) kN10

d) kN100

e) kN140

Answer: Alternative _______________ [1/0]

Why? Show the details of your calculations in the space provided below! [0/1]

Suggested solutions: Answer: Alternative b:14 kN

kNNmg 14917138sin100.1sin5 == [0/1]

2. In the diagram below, a N.30 force due north and a N.40 force due east act

concurrently (simultaneously) on an object, as shown in the diagram below. The

additional force necessary to bring the object into a

state of equilibrium is

a) N.10 at .45 south of west.

b) N.70 at .53 south of west.

c) N.50 at .37 south of west.

d) N.70 at .53 north of east.

e) N.50 at .37 north of east.

f) N.10 at .45 north of east.

g) =F Answer: Alternative _______________ [1/0]

Why? Show the details of your calculations in the space provided below! Draw necessary

figures. [1/1]

Suggested solutions: Answer: Alternative c:additional force at south of west.N.50 .37

The resultant force

is ( ) ( ) NNF .50.40.30 22 =+= [1/0]

at

= .37

40

30tan 1 north of east.

Therefore, as illustrated in the figure below

additional force at south of westnecessary to bring the object into a state of equilibrium. [0/1]

N.50 .37

N.30

N.40

N.30

N.40.37

.37

N.50


3/19



3. A cars performance is tested on various horizontal road surfaces. The brakes are applied,causing the rubber tires of the car to slide along the road without rolling. The tires

encounter the greatest force of friction to stop the car on

3. A cars performance is tested on various horizontal road surfaces. The brakes are applied,causing the rubber tires of the car to slide along the road without rolling. The tires

encounter the greatest force of friction to stop the car on

a) wet concretea) wet concreteb) dry concreteb) dry concretec) dry asphaltc) dry asphaltd) wet asphalt.d) wet asphalt.

Answer: Alternative _______________ [1/0]Answer: Alternative _______________ [1/0]

Answer: Alternative (b)

Base your answers to the questions 4 to 6on the

information and the diagram below. Note that the VG-

points are given to your explanations that give you theopportunity to demonstrate the depth of your

understandings of the physics of the problem. The

explanations must be written clearly in the space

provided.

4. The applied force on the object is represented by thevector:

a. A b. B c. C

d. D Answer: Alternative _______________ [1/0]

Identify each force A - D and explain in the space provided below why the applied

force is the one you chose above. [2/1]

Suggested solutions: Answer: Alternative (D): The applied force on theobject is represented by the vector D . This is due to the fact that frictionforce is always smaller than or equal to the applied force. ThereforeforceD can not be the friction force, and it must therefore be the external

force. Therefore the object must be accelerating up the incline, and theapplied force on the object is represented by the vectorD . [0/1]If the angle of the incline is

The vector A is the normal force which is cosmgFN = .

The vector B is the sum of the component of the weight parallel tothe surface of the incline and that of the friction, i.e.:

cossin + mgmg . [1/0]

The vector C is the component of the weight normal to the surface ofthe incline, i.e.: cosmg .

The vector D is the applied force explained above.[1/0]

A

B

C

D


4/19



5. The object in the figure above is5. The object in the figure above isa. at rest.a. at rest.

b. accelerating down the incline in the directionb. accelerating down the incline in the direction B .c. moving down the incline at a constant velocity.d. moving up the incline at a constant velocity.

e. accelerating up the incline in the direction D .f. depending on the magnitude of the friction force it could be at rest, moving down or

going up the incline.

Answer: Alternatives: ______________________ [1/0]

Why? Explain in the space provided below! [0/1]

Suggested solutions: Answer: Alternative e: The object is acceleratingup the incline in the directionD . It is due to the fact that, as mentionedearlier, the friction force is always smaller than the applied force. D is the

applied force and it is larger than the force A which is the sum of thefriction force and the component of the weight parallel to the surface ofthe incline:

cossin += mgmgD

amFnetr

r

= maBD =

Therefore, there is the net force on the object and according to Newtonssecond law of motion the object, therefore, will accelerate upwards.

[0/1]

6. The angle of the incline is decreased. The friction force on the objecta) increases as the angle of the incline decreases.

b) is constant and is independent of the angle of the incline.c) decreases as the angle of the incline decreases.d) first increases for a while, and then decreases.e) first decreases for a while, and then increases.

Answer: Alternative(s): ______________________ [1/0]

Why? Explain in the space provided below! [0/2]Suggested solutions: Answer: Alternative (a) The friction force onthe object increases as the angle of the incline is decreased. Thefriction force is part of the vector B , and due to the fact that theonly normal force is that of cosmgFN = and the friction force is

always proportional to the normal force, i.e. cosmgFf Nf == ,

and due to the fact that the cosine function is a decreasing functionof the angle in the first quadrant, 900 , the friction forceincreases as the angle of the incline is decreased. [0/2]


5/19



7. Tip of iceberg: The density of ice is 3/ whereas that for sea water is

3/ . What fraction of an iceberg is below the surface of the water? [1/2]

917.0 cmg

025.1 cmg

Suggested solutions: Answer: %5.89 of the volume of

the ice is beneath the surface of the water.

Data: ;3/917.0 cmgice =3/025.1 cmgSeaWater =

Due to the fact that the iceberg is in balance, thebuoyant force must be equal to the weight of theiceberg. Lets assume the total volume of the iceberg is

. Lets also assume that the total volume of

the part of the ice which is below the sea water is.

iTTotalice VV _

iBSWSurfaceTheBelowice VV ___

The buoyant force is the weight of the displacedseawater: gVgVF iBSWiBSWSeaWaterB == 025.1 .

The total weight of the iceberg is:gVgVmg iTiceiT == 917.0

Therefore:gVgVmgF iceiTiBSWSeaWaterB /=/=

iTSeaWater

iceiT

iTSeaWater

iBSWSeaWater

V

V

V

V

/

/=

/

/

%5.89895.0

025.1

917.0===

SeaWater

ice

iT

iBSW

V

V

Answer: %5.89 of the volume of the ice is beneath the surface of the

water.Second method (numerical)

gVgVmgF iTiBSWB /=/= 917.0025.1

%5.89895.08946.0917.0

025.1

917.0

025.1

025.1===

=

iTiT

iBSW

iT

iT

iT

iBSW

VV

V

V

V

V

V

m

gVF iBSWSeaWaterB =

gVmg iceiT =


6/19



8. Susanna stands on a digital bathroom scale in an elevator. The elevator does not have anywindow and is in general very quiet and is very difficult without hearing

the bell and seeing the floor-number to know if it is moving or not.

Susanna has designed an experiment which by using it, you can c

not only the magnitude of the acceleration of the elevator, but you cansay if it is accelerating up or down. Susannas mass is

8. Susanna stands on a digital bathroom scale in an elevator. The elevator does not have anywindow and is in general very quiet and is very difficult without hearing

the bell and seeing the floor-number to know if it is moving or not.

Susanna has designed an experiment which by using it, you can c

not only the magnitude of the acceleration of the elevator, but you cansay if it is accelerating up or down. Susannas mass is

alculatealculate

kg0.55 . The

scale reads kg0.51 .

a) The elevator accelerates upwards at the rate of 2/ sm 93.0

b) The elevator accelerates downwards at the rate of 2/0 sm .71.

c) The elevator moves downwards at constant velocity.

d) The elevator accelerates downwards at the rate of 2/0 sm .93.

e) The elevator accelerates upwards at the rate of 2/ sm 71.0.0

f) The elevator moves upwards at constant velocity.

Answer: Alternative(s): ____________________ [1/0]

Why? Draw a free-body diagram and explain your reasoning and

solutions in the space provided below! [1/1]

Suggested solutions: Answer: Alternative (b) [1/0]Due to the fact that Susannas weight is larger than herapparent weight (normal force NF ), we may realize that the

elevator is accelerating downwards as illustrated in the free-

body-figure to the right. Taking the direction of elevatorsacceleration (i.e. downwards) positive, Newtons second lawof motion may be expressed as:

maFmg N =

downwardssmm

Fg

m

Fmga NN 2/71.0

55

8.9518.9 =

==

= [0/1]

9. Suggest two different experimental methods to measure the static coefficient of friction.Explain in sufficient detail why the methods suggested may give reasonably correct result

for the coefficient of static friction. [1/1]

Suggested Answer: We may use an incline to obtain experimentally the static coefficient

of friction between the surfaces under the investigation:We may raise the angle of the incline (surface where the object is)gradually until the object starts moving. The tangent of the angle ofthe incline is that of the static friction: Cs tan=

C

C

CsCsC gmgm

tancos

sin

0cossin ===//// [0/1]

m

mg

NF

a


7/19



We may use a dynamometer to measure instantly the coefficient ofstatic and kinetic friction.We pull the object by a dynamometer on a horizontal surface until itstarts moving. The force measured by the dynamometer is exactly

equal to that of the static friction force: == mgFfsNsf mg

ffs=

The coefficient of the static friction is just the force measured by thedynamometer divided by the weight of the object. [1/0]

When assessing your work in the problems 10-12 I am going to consider: How well you have presented your work. How systematic and general your presentation, your solution and

your reasoning are. If your calculations are correct. Your analysis of the results and your conclusions.

How well you are using the mathematical and physical language inyour presentation.

10. A g0.130 necklace is suspected of being gold-plated lead instead of pure gold. If it is

dropped into a full glass of water g0.10 of water spills over. What proportion of the

necklace is pure gold? Density of pure gold is 3gold /0.19 cmg= , and that of lead is3

lead /0.11 cmg= . Density of water is3

water /0.1 cmg= . [2/4/M1-M5]

Suggested solutions: Answer: Only %5.36 of the bracelet is made of the

pure gold. The rest, i.e. %5.63 is made of lead.Data: gm 0.130necklace = , gmw 0.10= ,

30.10 cm=V

First method :

3

necklace

necklacenecklace /0.13

0.10

0.130cmg

V

m===

gmm goldlead 0.130=+ gmm goldlead = 0.130 [0/1]30.10 cmVV goldlead =+

30.10 cmmm

gold

gold

ead

lead=+ [0/1]

Using 3gold /0.19 cmg= ,3

lead /0.11 cmg= , and gmm goldlead = 130 :

0.101911

130=+

goldgold mm [0/1]

0.101911

11

11

130

19

19=+

goldgold

mm [0/1]

19110.10110.13019 =+ goldgold mm

209011192470 =+ goldgold mm


8/19



380209024708 ==goldm ggmgold 50.478

380== [1/0]

%5.36365.00.130

50.47

necklace

===g

g

m

mgold [1/0]

Answer: Only %5.36 of the bracelet is made of the pure gold. The rest, i.e.%5.63 is made of lead.

Second method:

Lets assume : goldVx , and leadVy . Therefore, we may construct a two

simultaneous equations system as:

=+

=+3

cm10isvolumetotal10

g130ismasstotal1301119

yx

yx [0/2]

To solve this equation system, we may multiply both sides of the secondequation by 11 and subtract the results from the first one so y is

eliminated:

=+

=+

1101111

1301119

yx

yx

1101301119 = xx [1/0]208 =x

35.2

8

20cmx == [0/1]

gxmgold 5.475.21919 === [1/0]

%5.36365.0130

5.47===

necklace

gold

m

m [0/1]

Answer: Only %5.36 of the bracelet is made of the pure gold. The rest, i.e.

%5.63 is made of lead.

Note that this is the percentage weight, while the percentage volume is:

%2525.010

5.2===

necklace

gold

V

V

It is customary to use the mass percentage rather than volumepercentage of gold in jewelry. Why?1

1 You pay for the amount of gold (i.e. its mass) used in the jewelry not for its volume!


9/19



Third method:

Lets assume : goldmx , and leadmy . Therefore, we may construct a two

simultaneous equations system as: total mass is 130 g

=+

=+

3cm10isvolumetotal10

1119

g130ismasstotal130yx

yx [0/2]

To solve this equation system, we may first simplify the second equation aby multiplying both sides of the equation by 2091119 = :

=+

=+

09021911

130

yx

yx

We may use elimination method to solve the problem. This may beachieved by multiplying the both sides of the first equation by 19 andsubtract the second equation from the results:

=+

==+

09021911

4702130191919

yx

yx

090247021119 = xx

gxx 5.478

3803808 === Answer: gx 5.47=

ggxy 5.825.47130130 === Answer: gy 5.82=

Answer: The necklace is made ofg130 gx 5.47= gold and gy 5.82=

lead. Therefore only %5.36=necklace

gold

m

mof the necklace is made of gold.

%5.36365.0130

5.47===

necklace

gold

m

m [0/1]

Answer: Only %5.36 of the bracelet is made of the pure gold. The rest, i.e.

%5.63 is made of lead.

Note that this is the percentage weight, while the percentage volume is:

%2525.0

10

5.2===

necklace

gold

V

V

It is customary to use the mass percentage rather than volumepercentage of gold in jewelry. Why?2

2 You pay for the amount of gold (i.e. its mass) used in the jewelry not for its volume!


10/19



11. A uniform ladder of mass kgm 0.10 and length11. A uniform ladder of mass kgm 0.10 and length= mL 0.6= mL 0.6=

leans at an angle against a frictionless wall. The coefficient of

static friction between the ground and the ladder is 40.0= .

a) Draw on the figure all forces applied on the ladder.Identify the forces, for example normal force,friction,Calculate the minimum angle at w

the ladder will not slip? [1/2/M1 M5]

If the ladder ma

hich

kes an angle of = .70 with the ground,culate the forces exerted on the ladder by the ground

2/2]

the

b) caland the wall. [

c) determine the minimum coefficient of friction at

base of the ladder if the ladder is not to slip when a kg0.75 person stands5

the way up the ladder. [1/2/M1 M5]

Note that

4of

( ) cos90sin = , and

costan =

solution,

sin

Suggested s:Data: m 10= mL 0.6= , 4kg0. .0= , ?min= , = 70 ,

?=GF , ?=WF kgm 75P 0.er = , L5

4

a) The condition of static equilibrium requires that

sum of all forces on the ladder is zero: = 0Fr

and

the total torque about any point on the ladder mustbe zero: = 0r

. i.e.:

= downup FF , =left FF , andright = clockwisenterClockwise

Lets name the forces as

cou Figure [0/1]

wF , Gx ,F GyF , mg , and gmPer as illustrated in the

figure to the right.

GyF : Normal force from th gr nd n e laddee ou o th r.

GxF : The horizontal force from the ground on the ladder. This force is the

friction force.

wF : Normal force from the ground on the ladder.

mg : Weight of the ladder.

g : Weight of Per.mPer

The conditions for static equilibrium requires that: The sum of all forces up is equal to the sum of all forces down:

[1/0]

nter-clockwise about the er o h th

NmgFGy .988.90.10 ===

The sum of all clock-wise torques is equal to the total torque coupoint where the ladd t uc es e ground:

( )( )

tan2sin2

cos

sin2

90sin

sin90sin2 =

=

=/=

/ mgmgmgFFLmg

LWW

[0/1]

L

L

WF

ff

GyF


11/19



The sum of all forces to the right is The sum of all forces to the right is equal to the sum of all forces toequal to the sum of all forces to

the left:the left:tan2

FF WGx .==mg

Gy 392.39.

On the other hand, the maximum friction force on the ground isff 9840.0max NNNmgF ==== . This is maximum

horizontal force between the base of the ladder and the ground

Therefore:calculated above.

== 11

tan2WGx mg

mgFF

=

===

514.02

tan2

tantan2

1

CC

C

mg

mg

[1/2/M1M5]

=max Gyf Ff

MVG- quality The student demonstrates the highest quality

in solving problem 11a by:

M1 evelops the problem, usesgeneral methods with problem solving.Formulates and d Uses free-body-diagram, generalize

the problem. Uses conditionsnecessary for the static equilibrium,i.e. 0= xF , 0= yF , and

clockwisecounterclockwise = . Calculates

minimum critical angle:

the

51C .

M5 The presentation is structured, andmathematical-physical language is correct.

well structured

solution with spec

Presentation of a

ially a correctmathematical language.

Second method regardingcalculation of torques:he sum of all clock-wise

ereround is

Ttorques about the point whthe ladder touches the gequal to the total torquecounter-clockwise about the

point:

2

xyw

LgmLF =

y

wL

gmF2

= xL

y

xw

L

Lgm F =

2

We may notice that

x

y

L

L=tan

tan

1=

y

x

L

L

tan2

=mg

FW tan2

==mg

FF WGx

L

WF

ff

GyF

gm

yL

xL


12/19



b)Similarly for = .70 and 40.0= :

The sum of all forces up is equal to the sum of all forces down:[1/0]

The sum of all clock-wise torques is equal to the total torque counter-clockwise about the point where the ladder touches the ground:

NmgFGy .988.90.10 ===

NFmg

FFLmgL

WWW 8.1770sin2

20sin98

70sin2

20sin70sin20sin

2=

=

=/=

/[0/2]

The sum of all forces to the right is equal to the sum of all forces tothe left: . Note that this friction force is less than the

maximu

NFFGxW 8.17=

m friction force, i.e.NNNmgFf Gyf 392.39.9840.0 ==== , otherwise, the ladder

would slip. [1/0]

NFFFGyGxG

6.99988.17 2222 +=+= NFG

100

=

= .807.79

8.17

98tantan 11

Gx

Gy

F

F .80

c) Similarly for = .70 , using the additional weight

kgmPer 0.75= , L5

4, ?min =

( ) NgmmgF perGy .8338.90.750.10 =+=+=

70sin20sin5

420sin

2/=

/+

/Wper FLgm

Lmg

L

70sin20sin10820sin

105 =+ Wper Fgmmg

20sin70sin10

85

+= g

mmF

per

W

NFW 23220sin8.970sin10

758105=

+=

278.0833

232232833

833

232minminmin

minmin

===

==

==

NN

NFf

NFf

Gyf

Wf

MVG- quality The student demonstrates the highest quality insolving problem 11c by:

M1 Formulates and develops theproblem, uses general methodswith problem solving.

Uses free-body-diagram, generalizes theproblem. Uses conditions necessary for thestatic equilibrium, i.e. 0= xF , , and0= yF

clockwisecounterclockwise =

M5 The presentation is structured,and mathematical-physicallanguage is correct.

Presentation of a well structured solution with

specially a correct mathematical language.

L

ff

GyF

gmPer

gm

L5

4


13/19



M

m

12. A kgM 00.1= wooden block on a plain inclined at = 54 is connected to a bucket by

a light cord running over a frictionless pulley. The empty

bucket is g.200 . It is partially filled with g.800 sand.

friction between the block andThe coefficient of static and

the surface of the block are 50.0=s and 30.0=k

respectively.

You are going to investigate the

w

possibility of sliding of the

ooden block down as well as up the incline.

]

st be removed from the bucket until the wooden

block starts sliding down the incline. Includ

c. Calculate the acceleration of the block and th in the cordas the block start sliding down the incline after removing the minimum amount of

sand from the bucket. (If you could not solve the part a above, for partial credit, you

may assume that sand is removed from the bucket.) [1/3/M1M2M5]

d. Inv ust be added to the partially filled bucket (i.e.

) until the wooden block starts sliding up the incline. Include free-body-

diagram for both the block and the buc [1/2]

e. Calculate the acceleration of the block and the magnitude of the tension in the cordas the block start sliding up the incline after adding a minimum amount of sand to

the bucket. (If you could not solve the part a above, for partial credit, you may

assume that sand is added to the partially filled bucket, and the total mass of

the bucket f d is now ) [1/3/M1M2M5]

a. Show that the system at the present condition is stationary. [2/0

b. Investigate, how much sand mue free-body-diagram for both the block

and the bucket containing sand. [1/2]

e magnitude of the tension

g.500

estigate, how much sand m

g.1000

ket containing sand.

g.200

illed with san g.2001


14/19



( )sinMg

M

TF

cosmgf =f

sandmm+

( ) gmm sand +

TF

Sugge d solution:Data:

stekgM 00.1= , kggmm bucket 200.0.200 == , , = 54 , 50.0=s , 30.0=k ,

kggmsand 800.0.800 ==

Problem: removed ?=sandm , ?=a , ?=TF ; added ?=sandm , ?=a , ?=TF

a, b, d :If we show that the block neither can slide down nor it can slide up at the

sent situation we have shown that the system is [Note that the solution in this part

pre stationary.answers also the question 12d]

friction force,

o

Lets first assume that the block intends to slide up the incline. If weshow that the minimum m required in this case is more than thesand

amount of sand in the bucket at the present, i.e. more thangmsand .800= , we have shown that the bucket can not slide up the

incline at present situation:

As the sand is added to the bucket, it gets heavier and heavier. The

ff gradually increases to its maximum value ( ) cosMgs

in the opposite direction of the intended motion. Just before, the lastgrain of the sand is added to the bucket, the total force on the woodenblock, as well as the total force on the bucket containing sand is zero,according to the Newtons first law:

( )

( ) ( )

=+

=+

0sincos

0

Ts

sandT

FMgMg

gmmF

( )

( ) ( ) ( )

+=+

+=

gmmMgMg

gmmF

sands

sandT

sincos

( ) ( )( ) ( ) /+=/+ gmmgM sands sincos ( ) ( )( )+=+ sincosssand Mmm ( ) ( )( ) mMm ssand += sincos

( ) ( )( ) 900.0200.054sin54cos50.000.1 =+=sandm

900.0=sandm is more than the amount of sand in the bucket at the

present. Therefore, the partially filled bucket at the present can not pulldown the block. At least gmm sandsand 100800900900 === sand must

be added to the bucket if the block is to slide up the incline.

[1/2 for 12d]


15/19



o Lets assume that the block intends to slide down the incline. If wehow that the maximum amount of san req red in t case is

less than the amount of sand in the bucket at the present, i.e. less.0 , w hav how e bucket can not slide up

line at present situation:

As the sand is removed from the bucket, it gets lighter and lighter. Thefriction force,

s d m ui hissand

than msand 80= e e s n that thg

the inc

ff gradually increases to its maximum value ( ) cosMgs

in the opposite direction of the intended motion. Just before, the lastgrain of the sand is removed from the bucket, the total force on thewooden block, as well as the total force on the bucket containing sand

is zero, according to the Newtons first law:( )

( ) ( )

+=

+=T cossin MgFM

gmmF

sT

sand

g( ) ( ) ( ) cossin gMgmmgM ssand /+/+=/ [1/0]

( ) ( ) cossin MmmM ssand ++= ( ) ( ) mMMm ssand = cossin ( ) ( ) gkgmsand 315315.02.054cos0.15.054sin0.1 ===

gmsand 315= is less than the amount of sand in the bucket. Therefore,

the block at the present can not pull the partially filled bucket.kggggmm sandsand 485.0485315800315 ==== sand must be removed

from the bucket if the block is to slide down the incline.

c) As demonstrated above gggmm sandsand 485315800315 ===

the block is to slide down thethe kinetic friction takes over, and

We may solve the problem for

sand

must be removed from the bucket ifincline. As the block starts movingtherefore the system accelerates.

kgg 315.0=

The total mass of the partially filled bucket is:

msand 315=

kggggm 515315200315 515.0==+=+

If we name the tension in the cord as TF and denote the acce

leration of

e syst m by , according to Newtons second lawth e of motion, anda

M

( ) cosMgf sf =

TF

( )sinMg

TF

sandmm+

( ) gmm sand +


16/19



noting that the friction force is changed tonoting that the friction force is changed to ( ) cosMgf kfk = in the

opposite direction of motion as illustrated in the figure below:

( ) ( )

=

=

MaMgMg

magF

sT

T

cossin

515.0

hich may be solved using the elim

F

ination method. The acceleratedystem is illustrated in the figure below.

ws

( ) ( )( ) ( ) MaaMggMg

MagFMg

agFk

T

T +==

=515.0cos515.0sin

cossin

515.0515.0

Mk

( ) ( )( ) ( ) aMgMM k += 515.0515.0cossin

Answer:( ) ( )( )

( )MMM

a k

+

g=

515.0

515.0cossin

A tween theccording the equation above the difference be ( )sinMg and( )cosMgF kT + accelerates the total mass of the system .0 [M2]

kgM+5 .51

( ) ( )( )( )

2/76.00.1515.0

515.054cos0.13.054sin0.1sm

ga =

+

= Answer: 2/76.0 sma =

( )agFagFaTT

gFT

+=+= 515.0515.0515.0515. = 0515.0

( ) ( ) NNFFag TTT 4.544.576.08.9515.0515.0F =+=+= N4.5 Answer: If we remove sand from the bucket

mo the system

iately accelerates at the rate of

FT

, the block will start

ving down the incline, the kinetic friction takes over, and

g315

immed2

/76.0 sma = down the incline. Thetension in the cord is N.FT 4.5

MVG- quality The student demonstrates MVG-

quality in solving 12a through

M1 Formulates and develops the problem, usesgeneral methods with problem solving.

Use of the free-body-diagram, and

solving Newtons equation of motion

for both the bucket and the block.

M2 Analyses and interprets the results, concludesand evaluates if they are reasonable.

Through interpretation of the results.


Using clear structure and

mathematical-physically correct

language.

TF

a

Ng515.0

kg515.0

M

TF

( )sinMg

a

( ) cosMgf kfk =


17/19



d The solution for 12d is included in the solut a asomewhat different solution is presented here:

d The solution for 12d is included in the solut a asomewhat different solution is presented here:

) ion for part 12

the block is to slide up the incline, the following condition must betisfied:

) ion for part 12

the block is to slide up the incline, the following condition must betisfied:

IfIfsasa( ) ( ) ( )( ) ( ) ( )( ) sincossincos +>+/+>/+ ssandssand MmmgMgm m

( ) ( )( ) mMmsa snd +> sincos ( ) ( )( ) 200.054sin. 54cos500.1 +>nd sam

kgmsand 900.0>

kgmkgm sandaddedsandadded 100.0800.0900.0 __>>

Therefore, at least 100 gram sand must be added to the bucket of sandfor ck up [1/2]the blo to slide the incline.

) As mentioned in the section d above if we add o the

bucket, increasing the total mass of th k

the system starts sliding up the incline and the kinetic friction takesss

sand 8.9100.1

e g100 sand t

e sand in the buc et to kg900.0

over. The excess force causes the system to accelerate. The total maof the bucket and sand is now kg100.1 .

( ) Ngmmkgmm sand 78.10100.1100.1 g ===+=+

T diagram for the block as w inione,

law of motion

he free-body- ell as the bucket is plottedtohe figure below. If we take the directbjects in the figure above, as positiv

of the acceleration of theaccording to Newtons second

( )sinMg

M

TF

( ) cosMgsf =f

sandmm+

( ) gmm sand +

TF


18/19



( )sinMg

M

TF

sandmm+

( ) gmm sand +

TF

a

a kg100.1

N78.10

( ) cosMgf kfk =

Figure [0/1]

( ) ( )( ) ( ) MaaMgMgg

MaMgMgF

aFgk

add

kT

T +==

=100.1cossin100.1

cossin

100.1100.1

( ) ( ) ( ) aMMgMgg k += 100.1cossin100.1 ( ) ( )

( )MMgMgg

a k

+

=

100.1

cossin100.1

According the equation above the difference between the total weight ofthe partially filled bucket, i.e. andN78.10 ( ) ( ) cossin MgMg k+

kgMaccelerates the total mass of the system +100.1 . [0/1/M1M2]

( ) ( )( )( )

22 /54.0/535.054cos0.13.054sin0.1100.1

smsmg

=000.1100.1 +

a

= 2/54.0 sma

( )agFagFaFg TTT === 100.1100.1100.1100.1100.1 ( ) ( ) NNFFagF TTT .1019.10535.08.9100.1100.1 == NFT 10

Answer: If we remove g315 sand from the bucket, the block will start

moving down the incline, the kinetic friction immediately takes over, and

the system accelerate rate of s at the

2

/73.0 sma=

down the incline.The tension in the cord is NFT .10 . [1/1/M5]

MVG- quality The student demonstrates MVG-

quality in solving 12b through


Use of the free-body-diagram, and

solving Newtons equation of motion

for both the bucket and the block.


Through interpretation of the results.

M5 The presentation is structured, and

mathematical-physical language is correct.

Using clear structure and

mathematical-physically correctlanguage.


19/19


MVG- quality 10 11a 11b 12c 12e


M2 Analyses and interprets the results, concludes

and evaluates if they are reasonable.M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.


MVG- quality



M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.


solution+test+fyanvc09++ch+3,+4,+8 4,+9 2 5,+10!1!6+vectors,+newton's+laws,+torque,+pressure

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