11th grade workbook answers - Chapter-1 (Force & Motion) - Newton's Laws of Motion grade work… · · 2016-02-2911th Grade Chapter-1 Newton’s Laws Motion Activity - 1.3.1 Applications

11th Grade Physics Workbook METU Development Foundation High School 1

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11th Grade Chapter-1 Newton’s Laws Motion Activity - 1.3.1 Applications of Newton’s Laws


1.3 - NEWTON’S LAWS OF MOTION Activity – 1.3.1

Applications of Newton’s laws of motion 1.

Three blocks of masses 5 kg, 3 kg and 2 kg are tied with ropes and pulled by a horizontal force of F=60 N. The coefficient of friction between the blocks and the surface is 0,2. Calculate the acceleration of the blocks. (g=10 m/s2.) 2.

Two blocks are connected by a string on an inclined plane as shown in the figure. When the system is released, tension of the rope is measured to be 24 N. Calculate the mass of m1. (g=10 m/s2. Pulleys are supposed to be weightless and frictionless.)

3.

The inclined planes given in the figure above are all frictionless. The masses of the objects are related as mK=mL<mm and β is greater than α. Compare the accelerations of the objects when they are released. 4.

The surfaces given in the figures above are all frictionless. The masses of objects X, Y and Z are m, 2m and 3m respectively. When the systems are released, the accelerations of the systems are a1 and a2. The tensions in the connecting ropes are T1 and T2. Calculate the ratio of accelerations and tensions. (Pulleys are supposed to be weightless and frictionless.)






3.



Take all the objects as a single body having mass of (5+3+2)=10 kg and draw a free body

diagram to show all the forces on a diagram.

W=100 N

F=60 Nx

y

Ff=100.0,2 Ff=20 N

Fnet=mtotal.a

(60-20)=10.a

a=4 m/s2

FN=100 N

FN2

T

60 N

30o

60.cos30o

60.sin30o

x

y

(60.0,5)-T=6.a

Fnet=m.a

FN1

T

10m

30o 10m.cos30o

10m.sin30o

x

y

T-(10m.0,5)=m.a

Fnet=m.a

30-24=6.a a=1 m/s224-5m=m

m=4 kg







3.








3.



FN

mg

kmg.cosa

mg.sina

x

y

mg.sink=m.a

Fnet=m.a

a=g.sink

Acceleration depends on the gravitational acceleration and the angle of inclined plane. As the angle gets bigger acceleration gets bigger. It

is independent on the mass.

So; aL>aK=aM

For Figure-I;

mg=(m+2m).a1

mg=3m.a1

a1=g/3

T1=2mg/3

For Figure-II;

mg=(m+3m).a2

mg=4m.a1

a2=g/4

T2=3mg/4

a1

a2

4

3=

T1

T2

8

9=and



5.

When the system given in Figure-I is released, object Y starts to move rightward with an acceleration of “a”. When the system given in Figure-II is pulled by a horizontal force of “F”, object Y starts to move leftward with an acceleration of “a”. What is the magnitude of force F in terms of “mg”? (All the friction forces and the weight of the pulleys are ignored.) 6.

The masses of the objects K, L and M are 2 kg, 1 kg and 2 kg respectively. The system given in the figure is released. After a while the rope connecting objects K and L breaks down. What can be said about the motion of object M? (All the friction forces and the weight of the pulley are ignored.)

7.

Object K is fired from the bottom of an inclined plane with an initial speed of “v”, it can reach point X then moves downward. The magnitude of its acceleration toward point X is “a1” and the magnitude of its acceleration when it moves downward is “a2”. The coefficient of friction is 0,5. What is the ratio of a1 to a2? (g=10 m/s2.) 8.

An object of mass “m” is stationary on a horizontal surface. Then a horizontal force of “F” is applied on the object. The acceleration of the object versus the magnitude of the applied force graph is given above. Assume that the magnitude of the maximum static friction force is equal to the magnitude of the kinetic friction force. Calculate the mass of the object and the coefficient of friction.


5.



7.



For Figure-I;

2mg=(m+2m).a

2mg=3m.a

a=2g/3

For Figure-II;

F-2mg=(m+2m).a

F-2mg=3ma

F=3ma+2mg

F=3m(2g/3)+2mg

F=2mg+2mg=4mg

Firstly, the total weight on the right side is greater. K and L start to move downward

and M starts to move upward with a constant acceleration.

When the rope between K and L breaks down, the net force on the system will be

zero. M will move at steady speed which is equal to the speed that it has while the rope

breaks down .



5.



7.




5.



7.



moving upward; Fnet=m.a

(-mg.sin37o)-(0,5.mg.cos37o)=m.a1

(-mg.sin37o)-(Ff)=m.a1

(-6)-(4)=a1 then a1=-10 m/s2

moving dowward; Fnet=m.a

(mg.sin37o)-(0,5.mg.cos37o)=m.a2

(mg.sin37o)-(Ff)=m.a2

(6)-(4)=a2 then a2=2 m/s2a1

a2

5=

Ff=12 N

Fnet=m.a

F-Ff=m.a

27-12=m.5

m=3 kg

Ff = µ.FN

12= µ.3.10 then µ=0,4



Inertial & non-inertial reference frames 1.

The coefficient of friction between the car and the box is 0,5. What would be the minimum magnitude of the acceleration of the system in order to keep the box stationary with respect to the car? (g=10 m/s2.) 2.

The inclined plane moves with a constant acceleration. The box of mass m is stationary with respect to the inclined plane. Express the acceleration of the inclined plane in terms of inclination angle α . (Ignore friction.)

3.

The elevator moves upward with a constant acceleration of 3 m/s2. What is the magnitude of the tension in the rope that connects the 2 kg object to the ceiling of the elevator? (g=10 m/s2.)

4.

The horizontal surface is frictionless and the coefficient of friction between 1 kg object and 4 kg object is 0,3. What is the maximum magnitude of the horizontal force F in order to keep the 1 kg object stationary with respect to 4 kg object? (g=10 m/s2.)






3.


4.


11th Grade Chapter-1 Newton’s Laws Motion Activity - 1.3.2 Inertial & non-inertial reference frames

mg

x

y

Ff

FNFfictitious=ma

Ff = µ.FN

Ffictitious= ma

Object is stationary; Ffictitious= FN and Ff =mg

µ.m.a =m.g

0,5.m.a=m.10 a=20 m/s2

FN

mg

y

x

ma mg.sinα mg.cosα ma.sinα ma.cosα

mg.sinα mg.cosα ma.sinα ma.cosα





Object is stationary;

m.g.sinα=m.a.cosα

a=g.tanα






3.


4.







3.


4.



mg

ma

T

object is in equilibrium;

T=mg+ma

T=(2.10)+(2.3)

T=26 N

10 N

x

y

FN=10 N

Ff=0,3.10=3 NFfictitious=ma=1.a=a

Ffictitious=Ff

a=3 m/s2

They move together.

F=(1+4).3=15 N


5.

The elevator moves upward with a constant acceleration of 2 m/s2. a) What is the

magnitude of the acceleration of object X with respect to the elevator?

b) What are the magnitudes of the accelerations of objects X and Y with respect to the ground?

(All the friction forces and the weight of the pulley are ignored and g=10 m/s2.)

6.

A block of mass 1 kg is placed on an inclined plane of mass 3 kg. Inclined plane is pushed by a horizontal force of 40 N as shown in the figure. Calculate the acceleration of the block with respect to the inclined plane. (Ignore friction and g=10 m/s2.)

7.

When the car is accelerating toward right with a constant acceleration of “a”, boxes A and B remain stationary with respect to the car. What is the magnitude of the acceleration of the car (a)? (All the friction forces and the weight of the pulley are ignored and g=10 m/s2.) 8.

The horizontal surface is frictionless and the coefficient of friction between 2 kg object and 3 kg object is “µ”. When the system given in the figure is released, all objects move together with maximum acceleration that they can move together. Calculate the coefficient of friction (µ). (Pulley is frictionless and its weight is ignored. g=10 m/s2.)


5.





6.


7.




20N

2a=4 N

T

30 N

3a=6 N

Ta)

motion

motion

T-24=2.a/ 36-T=3.a/

36-24=5.a/

a/=2,4 m/s2

b) aX=0,4 m/s2 (downward)

ay=4,4 m/s2 (upward)

FN

mg

y

x

ma

37o

37o

0,8.ma

0,6.ma

0,8.mg

0,6.mg

0,8.ma - 0,6.mg = m.a

F=m.asystem

40=4.asystem

asystem=10 m/s2

8 - 6 = 1.a

a = 2 m/s2 (upward)


5.





6.


7.




5.





6.


7.




20-T=0

For object A;For object B;

It is stationary

with respect to the car.

T=20 N

It is stationary

with respect to the car.

T=mA.asystem

20=4.asystem

asystem=5 m/s2

For the system of the objects;

Fnet=mtotal.asystem

5.10=(2+3+5). asystem

asystem=5 m/s2

For 2 kg object;

Ffictitious=Ffriction then m. asystem=µ.m.g

µ= (asystem/g)=(5)/(10)=0,5

Chapter-1 Newton’s Laws of Motion Additional Problems


1.3 - NEWTON’S LAWS OF MOTION Additional Problems

1.

The system given in the figure is frictionless. When the system is pulled by a vertical force F, the acceleration of the pulley and 2 kg object would be 2 m/s2 and 5 m/s2 respectively. Calculate the magnitude of the force F in newtons.

(Ignore friction, pulley is not rotating and g=10 m/s2.) 2.

The objects move with a constant acceleration of 3 m/s2 when a horizontal force F is applied on 6 kg object. The horizontal surface is frictionless and the coefficient of friction between 4 kg object and 6 kg object is 0,3. Calculate the magnitude of the horizontal force F in newtons. (Pulley is frictionless and its weight of the pulley is ignored. g=10 m/s2.)

3.

The systems given above are frictionless. The ratio of the tensions T1/T2 = 3/5 when the systems are moving. What is the ratio of the masses m1/m2? (Pulleys are frictionless and weights of the pulleys are ignored.) 4.

When a horizontal force F is applied on 2 kg object, 2 kg and 4 kg objects move together. The horizontal surface is frictionless and the coefficient of friction between 4 kg object and 2 kg object is 0,5. Calculate the maximum magnitude of the horizontal force F to move the objects together. (g=10 m/s2.)

For 4 kg object;

T-mg=m.a

T-10=1.6

T=16 N

For pulley

F-2T-mpulley.g=mpulley.apullley

F-32-5=0,5.2

F=38 N

T-Ff=m1.a

T-(0,3.40)=4.a

T=4a+12

For 6 kg object;

F-T-Ff=m2.a

F-T-12=6.a

F-24=(6.3)+12

F=54 NT=(4.3)+12=24 N



1.




3.



For 1 kg object;

11th Grade




1.




3.





1.




3.



For Figure-I;

m2.g=(m1+m2).a1

For Figure-II;

2m2.g=(m2+2m2).a2

T1=m1.a1

T2=m2.a2=2m2g/3

T1

T2

=

3

5=

m1.m2.g

(m1+m2)

2m2.g

3

2m2.g=3m2.a2

a2=2g/3

2.(m1+m2)

3.m1

m1

m2

=2

3

For 2 kg object;

F-Ff=2.a

F-(0,5.20)=2.a

F=2a+10

For 4 kg object;

Ff=4.a

10=4.a

a=2,5 m/s2

F=(2.2,5)+10

F=15 N

11th Grade



5.

When the wagon moves rightward with a constant acceleration of “a”, the ball inside the wagon is in equilibrium with respect to the wagon as shown in the figure. Calculate the magnitude of the acceleration of the wagon. 6.

The boy of mass 70 kg measures his weight as 490 N when the elevator moves downward. What is the magnitude of the acceleration of the elevator? Is the elevator speeding up or slowing down? (g=10 m/s2.)

7.

The elevator moves downward and decelerates at 3m/s2. An object of mass 4 kg is pulled by a horizontal force of F and it accelerates at 2 m/s2 with respect to the elevator. If the coefficient of friction between the mass and the elevator’s floor is 0,4 then what is the magnitude of the force F? (g=10 m/s2.)

8.

Three objects A, B and C are stationary on horizontal surfaces. Horizontal forces are applied to move the objects. The acceleration versus applied force graphs of the objects are given above. Answer the following questions. a) Compare the frictions forces on the

objects. b) Compare the masses of the objects. c) Compare the coefficients of friction of the

horizontal surfaces where B and C move. d) Compare the coefficients of friction of the

horizontal surfaces where A and B move.


5.



7.


8.




horizontal surfaces where A and B move. mg

x

y

Ffictitious=ma

T

T.sin37o

T.cos37o

37o

ma=T.sin37o

mg=T.cos37o

a=3g/4

mg

ma

FN

FN+ma=mg

490=700-(70.a)

a=3 m/s2

11th Grade



5.



7.


8.






5.



7.


8.





mg

x

y

Ffictitious=maF

FN

Ff

Ff=0,4.(4.10+4.3)

Ff=20,8 N

Fnet=m.a

F-Ff=m.a

F-20,8=4.2

F=28,8 N

a) C>B=A

b) A<B=C

c) B<C

d) A<B

By using the graphs; forces that are applied to start the motion of the objects are compared.

The reciprocal of the slope of acceleration versus time graph gives the mass of the object.

Masses of B and C are equal and the friction forces on them are related as C>B. The same relation must be

among the coefficient of frictions.

The friction forces on A and B are the same. Mass and the coefficient of friction are inversely proportional.

11th Grade

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11th grade workbook answers - Chapter-1 (Force & Motion) - Newton's Laws of Motion grade work… · · 2016-02-2911th Grade Chapter-1 Newton’s Laws Motion Activity - 1.3.1 Applications