SPH3U

Unit Test #2 – Forces

Name: ________________

Instructions:

1. Read each question carefully before answering.

2. Express all final answers with the correct number of significant digits.

3. Calculators may not be shared. No iPod or Smartphone calculators.

4. Good Luck !!

/ 5 B1.1 I can use appropriate terminology related to forces.

1. Which of the following best describes a normal force? a) a force that resists the motion or attempted motion of an object

b) a perpendicular force exerted by a surface on a an object in contact with the other

c) the pull between two objects at a distance d) the force in a string or rope when pulled 2. Which object, when stationary, has the least inertia? a) a desk b) a feather c) a computer d) a coffee mug

3. What is the value of acceleration due to gravity near the Earth’s surface? a) 9.8 N b) 9.8 m/s c) 9.8 m/s2 d) 9.8 km/h 4. A person drinks 100g of water. How much weight has that person gained in doing so ? a) 0.98 N b) 9.8 N c) 98 kg d) 9.8 kg

5. What is the acceleration of a block that has a mass of 24 kg and experiences a net force of 40 N? a) 1.9 m/s2 b) 0.9 m/s2

c) 1.7 m/s2 d) 1.5 m/s2

Test Score Parent/Guardian Signature / 63 %

:

:0 F=@lkD(9.84$

0

Fnettmaa

= Fen

= 40¥kg

SPH3U

/ 18 B1.3

I can use free-body diagrams and algebraic equations to analyse the net

force and acceleration of an object.

6. A large box is being moved across a floor. Draw FBDs of the situations described and explain any differences in the diagrams. Your FBDs must include all the forces acting on the box. Forces must be properly labelled. (6 marks)

a) A student pulls the box to the right using a piece of rope. b) A student pushes the box to the right.

7. Calculate the net force acting on a 5.0 kg box sliding across the floor accelerating at 4.0 m/s2 [backward]. What is causing the net force? (4 marks)

8. A dogsled team has 4 dogs that pull a person and a sled with a combined mass of 120kg. They start from rest and reach a speed of 35 km/h in 2.5 s.

a) Determine the average net force contributed by each dog. (5 marks)

b) If the force applied by each dog is 150 N determine the frictional force acting on the sled and the coefficient of kinetic friction between the sled and the snow. (3 marks)

Fn

Meet;fgH¥¥e D%←%±5£b

Fg

In a) a ropeis pulling .:#

in b) student '

is in contact with box :¥a

Fw → ForwardEet : ma

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Fisher=§ok$(4oYD

Fr#=Fk Fat . -2.0×10 'N. Fk iscaiusngnetforce

.

a)vi=om1s 35¥10,Y÷x}÷oaVf=35km/h

¥9.724At=25s

Vfifttaat Fnetima

9.724=42 .ss) =@k§⇐89nfya= 3.89 M$2

Fnet = 467NEACH DOG = 46744=117 N IRON

Fix.→t*•→FT¥g FN±t=FttI

467N =6ooN⇒,

Fk = 600N - 467N

Fk = 133N

: . Fe=BoNMi¥t=¥h5#←⇒=o "

SPH3U

/ 8 B1.6

I can state Newton’s Laws and use them to explain the effect of forces

acting on an object.

9. Which statement best describes Newton’s third law? a) For every action force, there is a reaction force of lesser magnitude and in the same direction

b) For every action force, there is a reaction force of equal magnitude but opposite in direction.

c) For every action force, there is a reaction force of equal magnitude and in the same direction

d) For every action force there is a reaction force of lesser magnitude but in the opposite direction.

10. You push on a frictionless cart with a force of 20 N. According to Newton’s 3rd Law, there is a reaction force of equal magnitude in the opposite direction, yet the net force is not zero and the cart accelerates. Why is this the case? a) The applied force, not the net force , determines acceleration. b) The action and reaction forces act on different objects. c) The action force overcomes the reaction force. d) The acceleration depends on the mass.

11. You have your seat belt on in the passenger seat of a car. The car suddenly accelerates. Which statement best describes your motion? a) your body will suddenly move forward with respect to the seat

b) your head will suddenly move forward with respect to your body.

c) both your body and your head will move forward at the same rate.

d) your head and your body will suddenly move backwards with respect to the seat.

12. Which of the following correctly states Newton’s second law? a) F=a/m b) mF=a c) F=ma d) F < ma 13. An ice skater pushes off the boards. Which of the following is the reaction force? a) the force of the skater pushing off the boards

b) the force of the boards pushing back on the skater

c) the force of gravity d) the force of friction from the ice

14. Austin stands on his longboard and pushes on a wall with a magnitude of 89 N. The total mass of Austin and the board is 65kg. Assume there is no friction. Calculate Austin’s acceleration. (3 marks)

0

O

0

o

o

The Wall pushes back

E with 89N.

%&→Fntt'89N Fnejma

*g oi=Fn¥

89465kgail .4m/s2

SPH3U

/ 32 B1.9 I can analyse and solve multi-step problems related to forces.

15. Two boxes of mass 30.0 kg and 10.0 kg are at rest side by side as shown in the diagram below. You apply a force of 3.0 x 102 N on the first box for 5.0 s, and they both slide across the floor. The larger box has a force of friction of 180 N, and the smaller box has a force of friction of 60.0 N.

a) Calculate the acceleration of the boxes during the 5.0 s you apply the force. b) Calculate the acceleration of the boxes after you stop pushing them. (Slowing down) c) Calculate how long it will take the boxes to stop moving. d) Calculate the total distance travelled by the boxes. Hint: There are 2 parts to their motion !

(12 marks total)

b)

NopgstnigInto F←##→Fa K

l¥y→Fm+ WFG

A) Fg FNEt=Fk=Ma

tI=18DNt6oN 240N Edwards],4oa

Fk=24oN →A=6mlj[back]

Fa . 300N 9-6%2

Fn*=FA- FK= 300N -240N

Fn*= 60N

oi=FntI=6dodsFmb2

10 Push for 5 seconds

����������� ������������������ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~��������������������������������� ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ2����������� ������������������ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~��������������������������������� ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ Boxes slow to a stop

01st5secad-

@Vi=7

.sn/svi=Om1sVgs0m/sVf=

?

aol.sn/s2a=-6m/s2at=5.0sAt

= ?

Vt=4taAtVt=§mk)( of Vfivitaat

Vf= 7.5mA Onls 'T

.sn/s.6mf.At6mlsat=7.5n1s1st=7.5m#b2

bt= 1.25 s

at =12s

d) H÷sdifdt . Kate

.tn#DHsXsI=esRHtssdEB

Dd= 18.75 m

LAST

1.25¥#Itza(t5ad = ¥f6nD¢25§ad = 4.69 M

Dofotd = 18.75Mt 469 m

Ddtotd = 23 m

SPH3U 16. In the diagram below a locomotive (6.4 x 105 kg) is used to pull two railway cars. The locomotive is attached to railway Car 1 (5.0 x 105kg), and railway car 1 is attached to railway car 2 (3.6 x 105kg) by the same locking mechanism. An engineer tests the mechanism and estimates that it can only withstand 2.0 x 107 N of force. The coefficient of kinetic friction between the wheels and the track is 0.42. Determine the maximum acceleration of the train that does not break either locking mechanism. (10 marks)

¥

iJnEtFnet-F-FkMa-Ft-Mxmgm-8.6xlJkgFT-2.oxio7N@6xioEDai2.oxkir-o4Dko6xDfD8.bx

105kg a = 2.0×107-35397605

8.6×10 a= 16 460 240

A = 16 460 240.

8.6×105As 1945

SPH3U 17. A 2.0 kg mass, placed on a smooth (no friction) , level table is attached by a light string passing over a frictionless pulley to a 5.0 kg mass hanging freely over the edge of the table, as shown in the diagram. Calculate the acceleration of the masses and the tension in the string. (10 marks)

F

:•}→¥n*net

Fg •|

FNEt=FT fbtrttmatt Fg

§okDa=FTFneima

FNETFGE

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¥i¥¥n¥⇐ImIiI±"Ft=14N #

a