Chapter 5 - Laws of Motion (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 5 - Laws of Motion (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
 w = Weight of plank,
 f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
 R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
        wall, 
N R ( ) = Reaction to N i.e. normal force on
             sphere B by sphere A,
    w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
 pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N 
Force acting on cylinder
Page 2


Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
 w = Weight of plank,
 f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
 R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
        wall, 
N R ( ) = Reaction to N i.e. normal force on
             sphere B by sphere A,
    w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
 pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N 
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
 
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
                            AB l =
    ( sin ) T l w
l
30
2
° =
Þ                    T w =
6. See figure (answer to question no. 3)
  sin q =
+
OA
OB BC
          =
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
      4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\          f = + ( ) 2.4 3 6
           =10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ     a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l 
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Page 3


Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
 w = Weight of plank,
 f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
 R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
        wall, 
N R ( ) = Reaction to N i.e. normal force on
             sphere B by sphere A,
    w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
 pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N 
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
 
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
                            AB l =
    ( sin ) T l w
l
30
2
° =
Þ                    T w =
6. See figure (answer to question no. 3)
  sin q =
+
OA
OB BC
          =
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
      4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\          f = + ( ) 2.4 3 6
           =10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ     a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l 
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
           =10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\     Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
    Net force on 2 kg block = ´ 2 a
              = ´ 2 10
             =20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ = 
Þ      a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
  = ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e.,    T mg ma cos sin f = + q …(i)
and       T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
   = + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
 )
   =
7
4
2 2
m g
or T
mg
=
7
2
  =5 7 N
Dividing Eq. (i) by Eq. (ii),
      tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
Page 4


Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
 w = Weight of plank,
 f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
 R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
        wall, 
N R ( ) = Reaction to N i.e. normal force on
             sphere B by sphere A,
    w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
 pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N 
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
 
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
                            AB l =
    ( sin ) T l w
l
30
2
° =
Þ                    T w =
6. See figure (answer to question no. 3)
  sin q =
+
OA
OB BC
          =
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
      4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\          f = + ( ) 2.4 3 6
           =10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ     a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l 
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
           =10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\     Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
    Net force on 2 kg block = ´ 2 a
              = ´ 2 10
             =20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ = 
Þ      a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
  = ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e.,    T mg ma cos sin f = + q …(i)
and       T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
   = + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
 )
   =
7
4
2 2
m g
or T
mg
=
7
2
  =5 7 N
Dividing Eq. (i) by Eq. (ii),
      tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
           =
+
a
g a
cos
sin
q
q
           =
+
cos
sin
q
q 2
           =
°
+ °
cos
sin
30
2 30
           =
3
5
i.e.,      f =
é
ë
ê
ù
û
ú
-
tan
1
3
5
6.
a =
+ +
8
2 1 1
 =2 m/s
2
Net force on 1 kg mass = - 8
2
T
\   8 1 2
2
- = ´ T
Þ      T
2
6 = N
Net force on 1 kg block = T
1
\     T a
1
2 = = ´ = 2 2 4 N 
Introductory Exercise 5.3
1. F g g = ° = 2 30 sin
For the system to remain at rest
T g
2
2 = …(i)
            T F T
2 1
+ = …(ii)
or          T g T
2 1
+ = …[ii (a)]
  T mg
1
= …(iii)
Substituting the values of T
1
 and T
2
 from
Eqs. (iii) and (i) in Eq. [ii(a)]
2g g mg + =
i.e.,     m =3 kg
2. As net downward force on the system is
zero, the system will be in equilibrium
\ T g
1
4 =
and T g
2
1 = 
\
T
T
1
2
4 =   
3. 2 2 g T a - =
Laws of Motion 77
2kg 1kg
F = 8 N
A B
2kg 1kg 8 N
T
1
T
1
T
2
T
2
B
A
T
1
T
1
mg
T
2
T
2
2 g
T
1
T
2
T
2
F
T
1
T
1
T
1
T
2
T
2
T
1
4g
1g
3g
T
T T
T
1g
2g
a
a
Page 5


Introductory Exercise 5.1
1. N = Nor mal force on cylinder by plank
R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N f cos q =
w N R + = sin q ,
N R ( ) = Reaction to N,
i.e.,normal force on plank by cylinder
R¢ = Normal force on plank by ground,
 w = Weight of plank,
 f ¢ = frictional force on plank by ground.
Resultant of f ¢ and R¢, N R ( ) and w pass
through point O.
2.
 R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N ¢ = Normal force on sphere A by sphere B,
w
A
= Weight of sphere A.
N R ¢ = cos q
N w N
A
¢ + = sin q
R¢ = Normal force on sphere B by right
        wall, 
N R ( ) = Reaction to N i.e. normal force on
             sphere B by sphere A,
    w
B
= Weight of sphere B,
R¢, N R ( ) and w
B
 pass through point O, the
centre sphere B.
3. N = Normal force on sphere by wall,
Laws of Motion 5
N(R)
Q
R'
q
O
w
f' P Ground
(Force acting on
plank)
N
q
R
w
A
N'
Force on sphere A
N (R)
q
R
w
B
O
R'
B
Force of sphere B
N
O
B
w
C
q
A
w
R
q
f
N 
Force acting on cylinder
w = Weight of sphere,
T = Tension in string.
4. Component of F
1
®
 
along x-axis : 4 30 2 3 cos ° = N
along y-axis : 4 30 2 sin ° = N
Component of F
2
®
along x-axis : 4 120 2 cos ° = - N
along y-axis : 4 120 2 3 sin ° = N
Component of F
3
®
along x-axis : 6 270 0 cos ° = N
along y-axis : 6 270 6 sin ° = - N
Component of F
4
®
along x-axis : 4 0 4 cos ° = N
along y-axis : 4 0 0 sin ° = N
5. Tak ing mo ment about point A
                            AB l =
    ( sin ) T l w
l
30
2
° =
Þ                    T w =
6. See figure (answer to question no. 3)
  sin q =
+
OA
OB BC
          =
+
=
a
a a
1
2
T w cos 30° =
or T w
3
2
=
or T w =
2
3
7. R f cos cos 30 3 60 ° + = °
i.e.,
R f 3
2
3
2
+ =
or R f 3 6 + = …(i)
and R f sin sin 30 60 10 ° + ° =
i.e., R f
1
2
3
2
10 + =
or R f + = 3 20 …(ii)
Substituting the value of f from Eq. (i) in
Eq. (ii)
R R + + = ( ) 3 6 3 20
      4 6 3 20 R + =
Þ R =
-
=
20 6 3
4
2.4 N
\          f = + ( ) 2.4 3 6
           =10.16 N
8.At point B (instantaneous vertical
acceleration only)
\ mg T ma - ° = sin 45 …(i)
At point A (instantaneous horizontal
acceleration only)
\ T ma cos 45° = …(ii)
Combining Eqs. (i) and (ii)
mg ma ma - =
Þ     a
g
=
2
Laws of Motion 75
30°
T sin 30°
B
O
w
A
T
AB = l 
30°
R
f
60°
60°
10 N
3N
N
A
T
T
45°
B
mg
Introductory Exercise 5.2
1. Acceleration of system
a =
+ + -
+ +
( ) ( ) 120 50
1 4 2
           =10 m/s
2
Let normal force between 1 kg block and
4 kg block = F
1
\     Net force on 1 kg block = - 120 N
\ a
F
=
- 120
1
1
or 10 120
1
= - F
i.e., F
1
110 = N
    Net force on 2 kg block = ´ 2 a
              = ´ 2 10
             =20 N
2. As, 4 30 2 30 g g sin sin ° > °
The normal force between the two blocks
will be zero.
3. N R
mg
( ) =
4
\ N
mg
=
4
As lift is moving downward with
acceleration a, the pseudo force on A will
be ma acting in the upward direction.
For the block to be at rest w.r.t. lift.
N ma mg + =
or
mg
ma mg
4
+ = 
Þ      a g =
3
4
4. Angle made by the string with the normal
to the ceiling = = ° q 30
As the train is moving with constant
velocity no pseudo force will act on the
plumb-bob.
Tension in spring = mg
  = ´ 1 10
= 10 N
5. Pseudo force ( ) = ma on plumb-bob will be
as shown in figure
T mg ma cos cos ( ) f = + ° - 90 q
i.e.,    T mg ma cos sin f = + q …(i)
and       T ma sin cos f = q
Squaring and adding Eqs. (i) and (ii),
T m g m a m ag
2 2 2 2 2 2 2
2 = + + sin sin q q
+ m a
2 2 2
cos q …(iii)
T m g m a m ag
2 2 2 2 2 2
= + + ( Qq = ° 30 )
   = + + × m g m
g
m
g
g
2 2 2
2
2
4 2
(Qa
g
=
2
 )
   =
7
4
2 2
m g
or T
mg
=
7
2
  =5 7 N
Dividing Eq. (i) by Eq. (ii),
      tan
cos
sin
f =
+
ma
mg ma
q
q
76 | Mechanics-1
mg
N (R) = N
ma
N
a
A
q = 30°
v (constant)
mg
T q
q = 30°
f
a
90° – q
ma
T
mg
           =
+
a
g a
cos
sin
q
q
           =
+
cos
sin
q
q 2
           =
°
+ °
cos
sin
30
2 30
           =
3
5
i.e.,      f =
é
ë
ê
ù
û
ú
-
tan
1
3
5
6.
a =
+ +
8
2 1 1
 =2 m/s
2
Net force on 1 kg mass = - 8
2
T
\   8 1 2
2
- = ´ T
Þ      T
2
6 = N
Net force on 1 kg block = T
1
\     T a
1
2 = = ´ = 2 2 4 N 
Introductory Exercise 5.3
1. F g g = ° = 2 30 sin
For the system to remain at rest
T g
2
2 = …(i)
            T F T
2 1
+ = …(ii)
or          T g T
2 1
+ = …[ii (a)]
  T mg
1
= …(iii)
Substituting the values of T
1
 and T
2
 from
Eqs. (iii) and (i) in Eq. [ii(a)]
2g g mg + =
i.e.,     m =3 kg
2. As net downward force on the system is
zero, the system will be in equilibrium
\ T g
1
4 =
and T g
2
1 = 
\
T
T
1
2
4 =   
3. 2 2 g T a - =
Laws of Motion 77
2kg 1kg
F = 8 N
A B
2kg 1kg 8 N
T
1
T
1
T
2
T
2
B
A
T
1
T
1
mg
T
2
T
2
2 g
T
1
T
2
T
2
F
T
1
T
1
T
1
T
2
T
2
T
1
4g
1g
3g
T
T T
T
1g
2g
a
a
T g a - = 1 1
Adding above two equations
1 3 g a =
\ a
g
=
3
Velocity of 1kg block 1 section after the
system is set in motion
            v at = + 0
              = ×
g
3
1
              =
g
3
 (upward)
On stopping 2 kg, the block of 1kg will go
upwards with retardation g. Time ( ) t¢
taken by the 1 kg block to attain zero
velocity will be given by the equation.
0
3
=
æ
è
ç
ö
ø
÷ + - ¢
g
g t ( )
Þ         t¢ =
1
3
 s 
If the 2 kg block is stopped just for a
moment (time being much-much less than 
1
3
 s), it will also start falling down when
the stopping time ends.
In t¢ =
æ
è
ç
ö
ø
÷
1
3
s time upward displacement of
1 kg block 
= = =
u
a
g
g
g
2 2
2
3
2 18
( / )
Downward displacement of 2 kg block
     = = ×
æ
è
ç
ö
ø
÷ =
1
2
1
2 3 18
2
2
at g
g g
As the two are just equal, the string will
again become taut after time 
1
3
 s.
4.        F g T a + - = 1 1       …(i)
and T g a - = 2 2 …(ii)
Adding Eqs. (i) and (ii),
F g a - = 1 3
\ a =
- 20 10
3
 =
10
3
 ms
-2
Introductory Exercise 5.4
1.           2 2 T a = ´             …(i)
and     1 2 g T a - = …(ii)
Solving Eqs. (i) and (ii),
a
g
=
3
\ Acceleration of 1 kg block 
2
2
3
20
3
a
g
= = ms
-2
Tension in the string
T
g
= =
3
10
3
 N
2.       Mg T Ma - =             …(i)
78 | Mechanics-1
T
T T
T
F
2g
a
1g
a
T T
2 kg
2T
a
1g
T
T
T T
2a
2T
T T
mg
T
T
a
M
T
T
a
Read More
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