Chapter 5 - Laws of Motion (Part - 2) - 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 - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


82 | Mechanics-1
6.
                  
T
W
2
2
= and
T
T
2
1
2
=
7.
N = 40 …(i)
f = 20 …(ii)
and f N x ´ = ´ 10 …(iii)
8.
T f N + ° = ° cos sin 30 30 …(i)
N f W cos sin 30 30 °+ ° = …(ii)
T R fR ´ = …(iii)
9.
V
T
W + =
3
2
…(i)
         H
T
=
2
…(ii)
Net moment about O = zero
\ . W
l T
l ´ = ´
2
3
2
…(iii)
10. (a) a =
-
+ +
=
100 40
6 4 10
3
2
m/s
(b) Net force = ma
\           F F
6 4
18 12 = = N, N and F
10
= 30 N
(c) N F - = = 40 30
10
\            N = 70  N.
 11. a
F
m m m
=
+ +
= =
1 2 3
60
60
1 m/s
2
(a)             T m a
1 1
10 = = N
          T T m a
2 1 2
- =
\        T
2
10 20 - =
\                T
2
30 = N
(b)  T
1
0 = . New acceleration 
a¢ = =
60
50
1.2 m s
2
/
T m a
2 2
24 = ¢ = N
12. (a) 
     T g a
1
2 2 - =
(b) 
           T g a
2
5 5 - =
x
f
10 cm
W = 20 N
N
40 N
N
T
30°
30°
f
W
W
V
O
T/2 H
3 T
2
1.9 kg
0.1 kg
T
1
2 g
a
45°
W
T
1
T
2
1.9 kg
0.2 kg
T
2
5 g
a
2.9 kg
Page 2


82 | Mechanics-1
6.
                  
T
W
2
2
= and
T
T
2
1
2
=
7.
N = 40 …(i)
f = 20 …(ii)
and f N x ´ = ´ 10 …(iii)
8.
T f N + ° = ° cos sin 30 30 …(i)
N f W cos sin 30 30 °+ ° = …(ii)
T R fR ´ = …(iii)
9.
V
T
W + =
3
2
…(i)
         H
T
=
2
…(ii)
Net moment about O = zero
\ . W
l T
l ´ = ´
2
3
2
…(iii)
10. (a) a =
-
+ +
=
100 40
6 4 10
3
2
m/s
(b) Net force = ma
\           F F
6 4
18 12 = = N, N and F
10
= 30 N
(c) N F - = = 40 30
10
\            N = 70  N.
 11. a
F
m m m
=
+ +
= =
1 2 3
60
60
1 m/s
2
(a)             T m a
1 1
10 = = N
          T T m a
2 1 2
- =
\        T
2
10 20 - =
\                T
2
30 = N
(b)  T
1
0 = . New acceleration 
a¢ = =
60
50
1.2 m s
2
/
T m a
2 2
24 = ¢ = N
12. (a) 
     T g a
1
2 2 - =
(b) 
           T g a
2
5 5 - =
x
f
10 cm
W = 20 N
N
40 N
N
T
30°
30°
f
W
W
V
O
T/2 H
3 T
2
1.9 kg
0.1 kg
T
1
2 g
a
45°
W
T
1
T
2
1.9 kg
0.2 kg
T
2
5 g
a
2.9 kg
13. (a) a
g
=
- 200 16
16
(b) T g a
1
11 11 - =
(c) T g a
2
9 9 - =
14. If the monkey exerts a force F on the rope
upwards, then same force F transfers to
bananas also. If monkey releases her hold
on rope both monkey and bananas fall
freely under gravity.
15. Tension on B T =
Tension on A T = 3
Now in these situations a
T
µ
1
16. x x x
A C B
+ + = 2 constant.
Differentiating twice w.r.t. time we get the 
acceleration relation.
17.
a
a
A
B
= sin q
\    a a
A B
= sin q
18. x y + = 6                  …(i)
y x - = 4 …(ii)
Solving, we get
           x = 1 m/s
2
19.                     a
g g
=
-
=
7 3
10
4 m/s
2
      40 4
1
- = T a
30 3
1 2
+ - = T T a
        T a
3
10 1 - =
20.
T a
1
1 = …(i)
T a a
r 1
20 2 2 - = - ( / ) …(ii)
30 3 2
1
- = + T a a
r
( / ) …(iii)
21.
T Mg Ma - ° = sin 30 …(i)
2 2 2
2
Mg T M
a
- = × …(ii)
22. T a = 1 …(i)
10 1 - = T a …(ii)
23.
50 2 5 - = T a …(i)
T a - = 40 4 2 ( ) …(ii)
24. (a) N = 40 N, m
s
N = 24 N
F N
s
< m
      \  f = 20 N and a = 0
(b) N = 20 N, m
s
N = 12 N
and m
k
N = 8 N
Laws of Motion 83
q
q
a
B
a
A
x
y
x
y
1
2
3
a
r
2
3
1
a
a
r
T
1
T
1
2T
1
T
1
a
2
T
1
T = T
1 2
T
a
2M
2T
2 Mg
a
2
M
Mg sin 30°
B
2T
50
a
A
2T
40
2a
Page 3


82 | Mechanics-1
6.
                  
T
W
2
2
= and
T
T
2
1
2
=
7.
N = 40 …(i)
f = 20 …(ii)
and f N x ´ = ´ 10 …(iii)
8.
T f N + ° = ° cos sin 30 30 …(i)
N f W cos sin 30 30 °+ ° = …(ii)
T R fR ´ = …(iii)
9.
V
T
W + =
3
2
…(i)
         H
T
=
2
…(ii)
Net moment about O = zero
\ . W
l T
l ´ = ´
2
3
2
…(iii)
10. (a) a =
-
+ +
=
100 40
6 4 10
3
2
m/s
(b) Net force = ma
\           F F
6 4
18 12 = = N, N and F
10
= 30 N
(c) N F - = = 40 30
10
\            N = 70  N.
 11. a
F
m m m
=
+ +
= =
1 2 3
60
60
1 m/s
2
(a)             T m a
1 1
10 = = N
          T T m a
2 1 2
- =
\        T
2
10 20 - =
\                T
2
30 = N
(b)  T
1
0 = . New acceleration 
a¢ = =
60
50
1.2 m s
2
/
T m a
2 2
24 = ¢ = N
12. (a) 
     T g a
1
2 2 - =
(b) 
           T g a
2
5 5 - =
x
f
10 cm
W = 20 N
N
40 N
N
T
30°
30°
f
W
W
V
O
T/2 H
3 T
2
1.9 kg
0.1 kg
T
1
2 g
a
45°
W
T
1
T
2
1.9 kg
0.2 kg
T
2
5 g
a
2.9 kg
13. (a) a
g
=
- 200 16
16
(b) T g a
1
11 11 - =
(c) T g a
2
9 9 - =
14. If the monkey exerts a force F on the rope
upwards, then same force F transfers to
bananas also. If monkey releases her hold
on rope both monkey and bananas fall
freely under gravity.
15. Tension on B T =
Tension on A T = 3
Now in these situations a
T
µ
1
16. x x x
A C B
+ + = 2 constant.
Differentiating twice w.r.t. time we get the 
acceleration relation.
17.
a
a
A
B
= sin q
\    a a
A B
= sin q
18. x y + = 6                  …(i)
y x - = 4 …(ii)
Solving, we get
           x = 1 m/s
2
19.                     a
g g
=
-
=
7 3
10
4 m/s
2
      40 4
1
- = T a
30 3
1 2
+ - = T T a
        T a
3
10 1 - =
20.
T a
1
1 = …(i)
T a a
r 1
20 2 2 - = - ( / ) …(ii)
30 3 2
1
- = + T a a
r
( / ) …(iii)
21.
T Mg Ma - ° = sin 30 …(i)
2 2 2
2
Mg T M
a
- = × …(ii)
22. T a = 1 …(i)
10 1 - = T a …(ii)
23.
50 2 5 - = T a …(i)
T a - = 40 4 2 ( ) …(ii)
24. (a) N = 40 N, m
s
N = 24 N
F N
s
< m
      \  f = 20 N and a = 0
(b) N = 20 N, m
s
N = 12 N
and m
k
N = 8 N
Laws of Motion 83
q
q
a
B
a
A
x
y
x
y
1
2
3
a
r
2
3
1
a
a
r
T
1
T
1
2T
1
T
1
a
2
T
1
T = T
1 2
T
a
2M
2T
2 Mg
a
2
M
Mg sin 30°
B
2T
50
a
A
2T
40
2a
F N
s
> m
     \              f N
k
= = m 8 N
     and a =
-
=
20 8
2
6 m/s
2
(c) N = - = 60 20 40 N
                m
s
N = 8 N and
                m
k
N = 4 N
Since, F N
s
cos 45° > m
\        f N
k
= = m 4 N
and    a =
- 20 4
6
             =
8
3
 m/s
2
25. a g = = m 3 m/s
2
(a) v at = 
    \   6 3 = t or t = 2 s
(b) s at =
1
2
2
 
\         s = ´ ´ =
1
2
3 4 6 m.
26. f = ´ ´ = 0 4 1 10 4 . N
a
f
1
1
4 = = m/s
2
a
f
2
2
2 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or 2 4 8 2 + = - t t
\     t = 1 s
(b)   v v
1 2
2 4 1 6 = = + ´ = m/s
(c)   s u t a t
1 1 1
2
1
2
= +
       s u t a t
2 2 2
2
1
2
= -
27. f = = ( . )( )( ) 06 2 10 12 N
      a
2
12
2
6 = = m/s
2
and a
1
12
1
12 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or u a t u a t
1 1 2 2
+ = +
or       3 6 18 12 - = - + t t
\             t =
7
6
 s        
(b) Common velocity at this instant is 
      v
1
 or v
2
.
(c) s u t a t
1 1 1
2
1
2
= + and
s u t a t
2 2 2
2
1
2
= +
28. N = 20 N
m
s
N = 16 N
and m
k
N = 12 N
Since, W = 20 N > m
s
N, friction m
k
N will
act.
\              a =
-
=
20 12
2
4 m/s
2
29. N = 20 N
mN = 16 N
Block will start moving when
F N = m
or 2 16 t =
or   t = 8 s.
After 8 s
a
t
t =
-
= -
2 16
2
8
i.e., a-t graph is a straight line with
positive slope and negative intercept.
30. N = 60 N, m
s
N = 36 N, m
k
N = 24 N
Block will start moving when
    F N
s
= m
84 | Mechanics-1
a
1
1 kg
2 kg
f
f
a
2
a
2
2 kg
f
a
1
f
1 kg
–ve +ve
Page 4


82 | Mechanics-1
6.
                  
T
W
2
2
= and
T
T
2
1
2
=
7.
N = 40 …(i)
f = 20 …(ii)
and f N x ´ = ´ 10 …(iii)
8.
T f N + ° = ° cos sin 30 30 …(i)
N f W cos sin 30 30 °+ ° = …(ii)
T R fR ´ = …(iii)
9.
V
T
W + =
3
2
…(i)
         H
T
=
2
…(ii)
Net moment about O = zero
\ . W
l T
l ´ = ´
2
3
2
…(iii)
10. (a) a =
-
+ +
=
100 40
6 4 10
3
2
m/s
(b) Net force = ma
\           F F
6 4
18 12 = = N, N and F
10
= 30 N
(c) N F - = = 40 30
10
\            N = 70  N.
 11. a
F
m m m
=
+ +
= =
1 2 3
60
60
1 m/s
2
(a)             T m a
1 1
10 = = N
          T T m a
2 1 2
- =
\        T
2
10 20 - =
\                T
2
30 = N
(b)  T
1
0 = . New acceleration 
a¢ = =
60
50
1.2 m s
2
/
T m a
2 2
24 = ¢ = N
12. (a) 
     T g a
1
2 2 - =
(b) 
           T g a
2
5 5 - =
x
f
10 cm
W = 20 N
N
40 N
N
T
30°
30°
f
W
W
V
O
T/2 H
3 T
2
1.9 kg
0.1 kg
T
1
2 g
a
45°
W
T
1
T
2
1.9 kg
0.2 kg
T
2
5 g
a
2.9 kg
13. (a) a
g
=
- 200 16
16
(b) T g a
1
11 11 - =
(c) T g a
2
9 9 - =
14. If the monkey exerts a force F on the rope
upwards, then same force F transfers to
bananas also. If monkey releases her hold
on rope both monkey and bananas fall
freely under gravity.
15. Tension on B T =
Tension on A T = 3
Now in these situations a
T
µ
1
16. x x x
A C B
+ + = 2 constant.
Differentiating twice w.r.t. time we get the 
acceleration relation.
17.
a
a
A
B
= sin q
\    a a
A B
= sin q
18. x y + = 6                  …(i)
y x - = 4 …(ii)
Solving, we get
           x = 1 m/s
2
19.                     a
g g
=
-
=
7 3
10
4 m/s
2
      40 4
1
- = T a
30 3
1 2
+ - = T T a
        T a
3
10 1 - =
20.
T a
1
1 = …(i)
T a a
r 1
20 2 2 - = - ( / ) …(ii)
30 3 2
1
- = + T a a
r
( / ) …(iii)
21.
T Mg Ma - ° = sin 30 …(i)
2 2 2
2
Mg T M
a
- = × …(ii)
22. T a = 1 …(i)
10 1 - = T a …(ii)
23.
50 2 5 - = T a …(i)
T a - = 40 4 2 ( ) …(ii)
24. (a) N = 40 N, m
s
N = 24 N
F N
s
< m
      \  f = 20 N and a = 0
(b) N = 20 N, m
s
N = 12 N
and m
k
N = 8 N
Laws of Motion 83
q
q
a
B
a
A
x
y
x
y
1
2
3
a
r
2
3
1
a
a
r
T
1
T
1
2T
1
T
1
a
2
T
1
T = T
1 2
T
a
2M
2T
2 Mg
a
2
M
Mg sin 30°
B
2T
50
a
A
2T
40
2a
F N
s
> m
     \              f N
k
= = m 8 N
     and a =
-
=
20 8
2
6 m/s
2
(c) N = - = 60 20 40 N
                m
s
N = 8 N and
                m
k
N = 4 N
Since, F N
s
cos 45° > m
\        f N
k
= = m 4 N
and    a =
- 20 4
6
             =
8
3
 m/s
2
25. a g = = m 3 m/s
2
(a) v at = 
    \   6 3 = t or t = 2 s
(b) s at =
1
2
2
 
\         s = ´ ´ =
1
2
3 4 6 m.
26. f = ´ ´ = 0 4 1 10 4 . N
a
f
1
1
4 = = m/s
2
a
f
2
2
2 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or 2 4 8 2 + = - t t
\     t = 1 s
(b)   v v
1 2
2 4 1 6 = = + ´ = m/s
(c)   s u t a t
1 1 1
2
1
2
= +
       s u t a t
2 2 2
2
1
2
= -
27. f = = ( . )( )( ) 06 2 10 12 N
      a
2
12
2
6 = = m/s
2
and a
1
12
1
12 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or u a t u a t
1 1 2 2
+ = +
or       3 6 18 12 - = - + t t
\             t =
7
6
 s        
(b) Common velocity at this instant is 
      v
1
 or v
2
.
(c) s u t a t
1 1 1
2
1
2
= + and
s u t a t
2 2 2
2
1
2
= +
28. N = 20 N
m
s
N = 16 N
and m
k
N = 12 N
Since, W = 20 N > m
s
N, friction m
k
N will
act.
\              a =
-
=
20 12
2
4 m/s
2
29. N = 20 N
mN = 16 N
Block will start moving when
F N = m
or 2 16 t =
or   t = 8 s.
After 8 s
a
t
t =
-
= -
2 16
2
8
i.e., a-t graph is a straight line with
positive slope and negative intercept.
30. N = 60 N, m
s
N = 36 N, m
k
N = 24 N
Block will start moving when
    F N
s
= m
84 | Mechanics-1
a
1
1 kg
2 kg
f
f
a
2
a
2
2 kg
f
a
1
f
1 kg
–ve +ve
or 4 36 t =
\    t = 9 s
After 9 s
a
t
t =
-
= -
4 24
6
2
3
4
31. N mg = = cos q 30 N
mg sin 30 30 3 ° = N »  52 N.
         m
s
N = 18 N and 
        m
k
N = 12 N
(a) 
              F = - = 52 18 34 N.
(b) 
F + = 12 52
        \           F = 40 N
(c) 
F - - = ´ 52 12 6 4
\            F = 88 N 
Objective Questions (Level-1)
Single Correct Option
1. a
mg F
m
g
F
m
=
-
= -
m m
A B
>
\ a a
A B
>
or ball A reaches earlier.
2. a
g g g
=
-
=
4 2
6 3
Now, 2 2
3
g T
g
- = 
\        T
g
= =
4
3
13 N
3.
3
2
100
1
T
= …(i)
T
T
1
2
2
= …(ii)
4.
mg T ma - =
\     a g
T
m
min
max
= -
= - = g
mg
m
g
2
3
3
5. a
g g g
=
-
=
10 5
15 3
10 10
3
g T
g
- = ´
\         T
g
= =
20
3
 Reading of spring
balance.
6. a
mg mg
m
=
°-
=
2 30
3
0
sin
\         T mg =
7. a g g g g
1
45 45 = - = °- ° sin cos sin cos q m q m
a g g
2
45 = ° sin sin q =
Now t
s
a
=
2
 or t
a
µ
1
\ 
t
t
a
a
1
2
2
1
=
or 2
45
45 45
=
°
°- °
g
g g
sin
sin cos m
Solving, we get m =
3
4
Laws of Motion 85
F
52 N
12 N
v
F
52 N
a
12 N
30°
T
1
T
2
100 N
mg
T
a
F
52 N
18 N
Page 5


82 | Mechanics-1
6.
                  
T
W
2
2
= and
T
T
2
1
2
=
7.
N = 40 …(i)
f = 20 …(ii)
and f N x ´ = ´ 10 …(iii)
8.
T f N + ° = ° cos sin 30 30 …(i)
N f W cos sin 30 30 °+ ° = …(ii)
T R fR ´ = …(iii)
9.
V
T
W + =
3
2
…(i)
         H
T
=
2
…(ii)
Net moment about O = zero
\ . W
l T
l ´ = ´
2
3
2
…(iii)
10. (a) a =
-
+ +
=
100 40
6 4 10
3
2
m/s
(b) Net force = ma
\           F F
6 4
18 12 = = N, N and F
10
= 30 N
(c) N F - = = 40 30
10
\            N = 70  N.
 11. a
F
m m m
=
+ +
= =
1 2 3
60
60
1 m/s
2
(a)             T m a
1 1
10 = = N
          T T m a
2 1 2
- =
\        T
2
10 20 - =
\                T
2
30 = N
(b)  T
1
0 = . New acceleration 
a¢ = =
60
50
1.2 m s
2
/
T m a
2 2
24 = ¢ = N
12. (a) 
     T g a
1
2 2 - =
(b) 
           T g a
2
5 5 - =
x
f
10 cm
W = 20 N
N
40 N
N
T
30°
30°
f
W
W
V
O
T/2 H
3 T
2
1.9 kg
0.1 kg
T
1
2 g
a
45°
W
T
1
T
2
1.9 kg
0.2 kg
T
2
5 g
a
2.9 kg
13. (a) a
g
=
- 200 16
16
(b) T g a
1
11 11 - =
(c) T g a
2
9 9 - =
14. If the monkey exerts a force F on the rope
upwards, then same force F transfers to
bananas also. If monkey releases her hold
on rope both monkey and bananas fall
freely under gravity.
15. Tension on B T =
Tension on A T = 3
Now in these situations a
T
µ
1
16. x x x
A C B
+ + = 2 constant.
Differentiating twice w.r.t. time we get the 
acceleration relation.
17.
a
a
A
B
= sin q
\    a a
A B
= sin q
18. x y + = 6                  …(i)
y x - = 4 …(ii)
Solving, we get
           x = 1 m/s
2
19.                     a
g g
=
-
=
7 3
10
4 m/s
2
      40 4
1
- = T a
30 3
1 2
+ - = T T a
        T a
3
10 1 - =
20.
T a
1
1 = …(i)
T a a
r 1
20 2 2 - = - ( / ) …(ii)
30 3 2
1
- = + T a a
r
( / ) …(iii)
21.
T Mg Ma - ° = sin 30 …(i)
2 2 2
2
Mg T M
a
- = × …(ii)
22. T a = 1 …(i)
10 1 - = T a …(ii)
23.
50 2 5 - = T a …(i)
T a - = 40 4 2 ( ) …(ii)
24. (a) N = 40 N, m
s
N = 24 N
F N
s
< m
      \  f = 20 N and a = 0
(b) N = 20 N, m
s
N = 12 N
and m
k
N = 8 N
Laws of Motion 83
q
q
a
B
a
A
x
y
x
y
1
2
3
a
r
2
3
1
a
a
r
T
1
T
1
2T
1
T
1
a
2
T
1
T = T
1 2
T
a
2M
2T
2 Mg
a
2
M
Mg sin 30°
B
2T
50
a
A
2T
40
2a
F N
s
> m
     \              f N
k
= = m 8 N
     and a =
-
=
20 8
2
6 m/s
2
(c) N = - = 60 20 40 N
                m
s
N = 8 N and
                m
k
N = 4 N
Since, F N
s
cos 45° > m
\        f N
k
= = m 4 N
and    a =
- 20 4
6
             =
8
3
 m/s
2
25. a g = = m 3 m/s
2
(a) v at = 
    \   6 3 = t or t = 2 s
(b) s at =
1
2
2
 
\         s = ´ ´ =
1
2
3 4 6 m.
26. f = ´ ´ = 0 4 1 10 4 . N
a
f
1
1
4 = = m/s
2
a
f
2
2
2 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or 2 4 8 2 + = - t t
\     t = 1 s
(b)   v v
1 2
2 4 1 6 = = + ´ = m/s
(c)   s u t a t
1 1 1
2
1
2
= +
       s u t a t
2 2 2
2
1
2
= -
27. f = = ( . )( )( ) 06 2 10 12 N
      a
2
12
2
6 = = m/s
2
and a
1
12
1
12 = = m/s
2
(a) Relative motion will stop when 
v v
1 2
=
or u a t u a t
1 1 2 2
+ = +
or       3 6 18 12 - = - + t t
\             t =
7
6
 s        
(b) Common velocity at this instant is 
      v
1
 or v
2
.
(c) s u t a t
1 1 1
2
1
2
= + and
s u t a t
2 2 2
2
1
2
= +
28. N = 20 N
m
s
N = 16 N
and m
k
N = 12 N
Since, W = 20 N > m
s
N, friction m
k
N will
act.
\              a =
-
=
20 12
2
4 m/s
2
29. N = 20 N
mN = 16 N
Block will start moving when
F N = m
or 2 16 t =
or   t = 8 s.
After 8 s
a
t
t =
-
= -
2 16
2
8
i.e., a-t graph is a straight line with
positive slope and negative intercept.
30. N = 60 N, m
s
N = 36 N, m
k
N = 24 N
Block will start moving when
    F N
s
= m
84 | Mechanics-1
a
1
1 kg
2 kg
f
f
a
2
a
2
2 kg
f
a
1
f
1 kg
–ve +ve
or 4 36 t =
\    t = 9 s
After 9 s
a
t
t =
-
= -
4 24
6
2
3
4
31. N mg = = cos q 30 N
mg sin 30 30 3 ° = N »  52 N.
         m
s
N = 18 N and 
        m
k
N = 12 N
(a) 
              F = - = 52 18 34 N.
(b) 
F + = 12 52
        \           F = 40 N
(c) 
F - - = ´ 52 12 6 4
\            F = 88 N 
Objective Questions (Level-1)
Single Correct Option
1. a
mg F
m
g
F
m
=
-
= -
m m
A B
>
\ a a
A B
>
or ball A reaches earlier.
2. a
g g g
=
-
=
4 2
6 3
Now, 2 2
3
g T
g
- = 
\        T
g
= =
4
3
13 N
3.
3
2
100
1
T
= …(i)
T
T
1
2
2
= …(ii)
4.
mg T ma - =
\     a g
T
m
min
max
= -
= - = g
mg
m
g
2
3
3
5. a
g g g
=
-
=
10 5
15 3
10 10
3
g T
g
- = ´
\         T
g
= =
20
3
 Reading of spring
balance.
6. a
mg mg
m
=
°-
=
2 30
3
0
sin
\         T mg =
7. a g g g g
1
45 45 = - = °- ° sin cos sin cos q m q m
a g g
2
45 = ° sin sin q =
Now t
s
a
=
2
 or t
a
µ
1
\ 
t
t
a
a
1
2
2
1
=
or 2
45
45 45
=
°
°- °
g
g g
sin
sin cos m
Solving, we get m =
3
4
Laws of Motion 85
F
52 N
12 N
v
F
52 N
a
12 N
30°
T
1
T
2
100 N
mg
T
a
F
52 N
18 N
8. F mg mg
1
= sin cos q + m q
F mg mg
2
= sin cos q - m q
Given that F F
1 2
2 =
9. For equilibrium of block, net force from
plane should be equal and opposite of
weight.
10. No solution is required.
11. Angle of repose q m = = °
-
tan ( )
1
30
\     h R R R = - = -
æ
è
ç
ö
ø
÷ cos q 1
3
2
12. Net pulling force = - = = 15 5 10 g g g F
Net retarding force = = = ( )( ) 0.2 5g g f
\                             a
F f
g =
-
=
25
9
25
     T g a g
1
5 5
9
5
- = = 
\           T g
1
34
5
=
    15 15
27
5
2
g T a g - = =
\            T
2
48
5
= g
\              
T
T
1
2
17
24
=
13.
Relative to lift, a g a
r
= + ( ) sin q along the 
                                         plane.
Now, t
s
a
r
=
2
             =
+
2L
g a ( ) sin q
14. f mg = sin q    (if block is at rest)
15. 2 30 T F cos ° =
 \          T
F
=
3
              a
T
m
=
° cos 60
                  =
F
m 2 3
16.
  N mg F = - sin q
m f q N N mg F = = f - (tan ) (tan )( sin )
F N cos q m =
17. mmg = ´ ´ = 0.2 4 10 8 N
At t = 2 s, F = 4 N
Since F mg < m
\     Force of friction f F = = 4 N
18. a g
1
= sin q
a g g
2
= sin cos q - m q
t
a
=
25
 or t
a
µ
1
t
t
a
a
1
2
2
1
=
1
2
=
- g g
g
sin cos
sin
q m q
q
19. Net accelaration of man relative to ground 
= + = a a a 2
T mg m a - = ( ) 2
\              T m g a = + ( ) 2
20.   a = 0
3 50 25 75 F g g = + = ( )
\ F = 25 g = 250 N
86 | Mechanics-1
q
a
v
q
F
h
q
T 60°
30°
F
a
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