HC Verma Solutions: Chapter 6 - Friction | Physics Class 11 - NEET PDF Download

 Page 1


6.1
SOLUTIONS TO CONCEPTS 
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0 ? R = mg ...(1)
Again ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 ?
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0 ? R = mg ...(1)
ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S = 
a 2
u v
2 2
?
= 
) 1 ( 2
10 0
2
?
?
= 
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f ? frictional force
F ? Applied force
From grap it can be seen that when applied force is zero, 
frictional force is zero.
4. From the free body diagram,
R – mg cos ? = 0 ? R = mg cos ? ..(1)
For the block
U = 0, s = 8m, t = 2sec.
?s = ut + ½ at
2
? 8 = 0 + ½ a 2
2
? a = 4m/s
2
Again, ?R + ma – mg sin ? = 0
? ? mg cos ? + ma – mg sin ? = 0 [from (1)]
? m( ?g cos ? + a – g sin ?) = 0
? ? × 10 × cos 30° = g sin 30° – a
??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4
? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11
? Co-efficient of kinetic friction between the two is 0.11.
5. From the free body diagram
4 – 4a – ?R + 4g sin 30° = 0 …(1) 
R – 4g cos 30° = 0 ...(2)
? R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0
? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2 
? s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma
?R
a
R
mg
ma
?R
a velocity
F
p
o
mg
R
30°
R
???
?R
mg
ma
4kg
30°
4N
R
???
?R
mg
ma
a velocity
Page 2


6.1
SOLUTIONS TO CONCEPTS 
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0 ? R = mg ...(1)
Again ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 ?
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0 ? R = mg ...(1)
ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S = 
a 2
u v
2 2
?
= 
) 1 ( 2
10 0
2
?
?
= 
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f ? frictional force
F ? Applied force
From grap it can be seen that when applied force is zero, 
frictional force is zero.
4. From the free body diagram,
R – mg cos ? = 0 ? R = mg cos ? ..(1)
For the block
U = 0, s = 8m, t = 2sec.
?s = ut + ½ at
2
? 8 = 0 + ½ a 2
2
? a = 4m/s
2
Again, ?R + ma – mg sin ? = 0
? ? mg cos ? + ma – mg sin ? = 0 [from (1)]
? m( ?g cos ? + a – g sin ?) = 0
? ? × 10 × cos 30° = g sin 30° – a
??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4
? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11
? Co-efficient of kinetic friction between the two is 0.11.
5. From the free body diagram
4 – 4a – ?R + 4g sin 30° = 0 …(1) 
R – 4g cos 30° = 0 ...(2)
? R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0
? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2 
? s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma
?R
a
R
mg
ma
?R
a velocity
F
p
o
mg
R
30°
R
???
?R
mg
ma
4kg
30°
4N
R
???
?R
mg
ma
a velocity
Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to 
the net force acting down the incline = ? R + 2 g sin 30°
= 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net 
force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – ?R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
? Force required is zero.
7. From the free body diagram
g = 10m/s
2
, m = 2kg, ? = 30°, ? = 0.2
R – mg cos ? - F sin ? = 0
? R = mg cos ? + F sin ?? ...(1)
And mg sin ? + ?R – F cos ? = 0
? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0
? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0
? F =
) cos sin (
) cos mg sin mg (
? ? ? ?
? ? ? ?
? F = 
) 2 / 3 ( ) 2 / 1 ( 2 . 0
) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2
? ?
? ? ? ? ? ?
= 
76 . 0
464 . 13
= 17.7N ? 17.5N ?
8. m ? mass of child
R – mg cos 45° = 0
? R = mg cos 45° = mg /v
2
...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - ?R
= mg sin 45° - ? mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration  = 
mass
Force
= 
m
) 2 2 ( m
= 2 2 m/s
2
?
9. Suppose, the body is accelerating down with acceleration ‘a’.
From the free body diagram
R – mg cos ? = 0
? R = mg cos ? ...(1)
ma + mg sin ? – ? R = 0
? a = 
m
) cos (sin mg ? ? ? ?
= g (sin ? – ? cos ?)
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at
2
? 0.5 = 0 + ½ a (0/5)
2
? a = 4m/s
2
...(2)
For the next half metre 
u` = 2m/s, a = 4m/s
2
, s= 0.5.
? 0.5 = 2t + (1/2) 4 t
2
? 2 t
2
+ 2 t – 0.5 =0
(body moving us)
?R
R
???
F
mg
(body moving down)
?R
R
???
mg
R
45°
?R
mg
F
30°
?R
R
30°
mg
mg
ma
?R
R
Page 3


6.1
SOLUTIONS TO CONCEPTS 
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0 ? R = mg ...(1)
Again ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 ?
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0 ? R = mg ...(1)
ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S = 
a 2
u v
2 2
?
= 
) 1 ( 2
10 0
2
?
?
= 
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f ? frictional force
F ? Applied force
From grap it can be seen that when applied force is zero, 
frictional force is zero.
4. From the free body diagram,
R – mg cos ? = 0 ? R = mg cos ? ..(1)
For the block
U = 0, s = 8m, t = 2sec.
?s = ut + ½ at
2
? 8 = 0 + ½ a 2
2
? a = 4m/s
2
Again, ?R + ma – mg sin ? = 0
? ? mg cos ? + ma – mg sin ? = 0 [from (1)]
? m( ?g cos ? + a – g sin ?) = 0
? ? × 10 × cos 30° = g sin 30° – a
??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4
? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11
? Co-efficient of kinetic friction between the two is 0.11.
5. From the free body diagram
4 – 4a – ?R + 4g sin 30° = 0 …(1) 
R – 4g cos 30° = 0 ...(2)
? R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0
? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2 
? s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma
?R
a
R
mg
ma
?R
a velocity
F
p
o
mg
R
30°
R
???
?R
mg
ma
4kg
30°
4N
R
???
?R
mg
ma
a velocity
Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to 
the net force acting down the incline = ? R + 2 g sin 30°
= 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net 
force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – ?R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
? Force required is zero.
7. From the free body diagram
g = 10m/s
2
, m = 2kg, ? = 30°, ? = 0.2
R – mg cos ? - F sin ? = 0
? R = mg cos ? + F sin ?? ...(1)
And mg sin ? + ?R – F cos ? = 0
? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0
? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0
? F =
) cos sin (
) cos mg sin mg (
? ? ? ?
? ? ? ?
? F = 
) 2 / 3 ( ) 2 / 1 ( 2 . 0
) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2
? ?
? ? ? ? ? ?
= 
76 . 0
464 . 13
= 17.7N ? 17.5N ?
8. m ? mass of child
R – mg cos 45° = 0
? R = mg cos 45° = mg /v
2
...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - ?R
= mg sin 45° - ? mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration  = 
mass
Force
= 
m
) 2 2 ( m
= 2 2 m/s
2
?
9. Suppose, the body is accelerating down with acceleration ‘a’.
From the free body diagram
R – mg cos ? = 0
? R = mg cos ? ...(1)
ma + mg sin ? – ? R = 0
? a = 
m
) cos (sin mg ? ? ? ?
= g (sin ? – ? cos ?)
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at
2
? 0.5 = 0 + ½ a (0/5)
2
? a = 4m/s
2
...(2)
For the next half metre 
u` = 2m/s, a = 4m/s
2
, s= 0.5.
? 0.5 = 2t + (1/2) 4 t
2
? 2 t
2
+ 2 t – 0.5 =0
(body moving us)
?R
R
???
F
mg
(body moving down)
?R
R
???
mg
R
45°
?R
mg
F
30°
?R
R
30°
mg
mg
ma
?R
R
Chapter 6
6.3
? 4 t
2
+ 4 t – 1 = 0
? = 
4 2
16 16 4
?
? ? ?
= 
8
656 . 1
= 0.207sec
Time taken to cover next half meter is 0.21sec.
10. f ? applied force
F
i
? contact force
F ? frictional force
R ? normal reaction
? = tan ? = F/R
When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction 
increase upto limiting friction ( ?R)
Before reaching limiting friction
F < ?R
? tan ? = 
R
R
R
F ?
? ? tan ? ? ? ? ??? tan
–1
? ?
11. From the free body diagram
T + 0.5a – 0.5 g = 0 ...(1)
?R + 1a + T
1
– T = 0 ...(2)
?R + 1a – T
1
= 0
?R + 1a = T
1
...(3)
From (2) & (3) ? ?R + a =  T – T
1
? T – T
1
= T
1
? T = 2T
1
Equation (2) becomes ?R + a + T
1
– 2T
1
= 0
? ?R + a – T
1
= 0
? T
1
= ?R + a = 0.2g + a ...(4)
Equation (1) becomes 2T
1
+ 0/5a – 0.5g = 0
? T
1
= 
2
a 5 . 0 g 5 . 0 ?
= 0.25g – 0.25a ...(5)
From (4) & (5)  0.2g + a = 0.25g – 0.25a 
? a = 
25 . 1
05 . 0
× 10 = 0.04 I 10 = 0.4m/s
2
a) Accln of 1kg blocks each is 0.4m/s
2
b) Tension T
1
= 0.2g + a + 0.4 = 2.4N
c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N
12. From the free body diagram
?
1
R + 1 – 16 = 0
? ?
1
(2g) + (–15) = 0
? ?
1
= 15/20 = 0.75
?
2
R
1
+ 4 × 0.5 + 16 – 4g sin 30° = 0
??
2
(20 3 ) + 2 + 16  – 20 = 0
? ?
2
= 
3 20
2
= 
32 . 17
1
= 0.057 ? 0.06
?Co-efficient of friction ?
1
= 0.75 & ?
2
= 0.06 ?
F
f
? ?
Fi
R
Limiting
Friction
?R
T 1
R
A  1g
1a
?
?R
R
1g
1a
?
0.5g
0.5g
a
B A
?=0.2
0.5kg
1kg 1kg
?=0.2
2×0.5
16N
?R 1
4g
16N=T
? 2R
4×0.5
30°
a
? 1
0.5 m/s
2
2kg
4kg
? 2
Page 4


6.1
SOLUTIONS TO CONCEPTS 
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0 ? R = mg ...(1)
Again ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 ?
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0 ? R = mg ...(1)
ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S = 
a 2
u v
2 2
?
= 
) 1 ( 2
10 0
2
?
?
= 
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f ? frictional force
F ? Applied force
From grap it can be seen that when applied force is zero, 
frictional force is zero.
4. From the free body diagram,
R – mg cos ? = 0 ? R = mg cos ? ..(1)
For the block
U = 0, s = 8m, t = 2sec.
?s = ut + ½ at
2
? 8 = 0 + ½ a 2
2
? a = 4m/s
2
Again, ?R + ma – mg sin ? = 0
? ? mg cos ? + ma – mg sin ? = 0 [from (1)]
? m( ?g cos ? + a – g sin ?) = 0
? ? × 10 × cos 30° = g sin 30° – a
??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4
? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11
? Co-efficient of kinetic friction between the two is 0.11.
5. From the free body diagram
4 – 4a – ?R + 4g sin 30° = 0 …(1) 
R – 4g cos 30° = 0 ...(2)
? R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0
? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2 
? s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma
?R
a
R
mg
ma
?R
a velocity
F
p
o
mg
R
30°
R
???
?R
mg
ma
4kg
30°
4N
R
???
?R
mg
ma
a velocity
Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to 
the net force acting down the incline = ? R + 2 g sin 30°
= 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net 
force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – ?R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
? Force required is zero.
7. From the free body diagram
g = 10m/s
2
, m = 2kg, ? = 30°, ? = 0.2
R – mg cos ? - F sin ? = 0
? R = mg cos ? + F sin ?? ...(1)
And mg sin ? + ?R – F cos ? = 0
? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0
? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0
? F =
) cos sin (
) cos mg sin mg (
? ? ? ?
? ? ? ?
? F = 
) 2 / 3 ( ) 2 / 1 ( 2 . 0
) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2
? ?
? ? ? ? ? ?
= 
76 . 0
464 . 13
= 17.7N ? 17.5N ?
8. m ? mass of child
R – mg cos 45° = 0
? R = mg cos 45° = mg /v
2
...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - ?R
= mg sin 45° - ? mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration  = 
mass
Force
= 
m
) 2 2 ( m
= 2 2 m/s
2
?
9. Suppose, the body is accelerating down with acceleration ‘a’.
From the free body diagram
R – mg cos ? = 0
? R = mg cos ? ...(1)
ma + mg sin ? – ? R = 0
? a = 
m
) cos (sin mg ? ? ? ?
= g (sin ? – ? cos ?)
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at
2
? 0.5 = 0 + ½ a (0/5)
2
? a = 4m/s
2
...(2)
For the next half metre 
u` = 2m/s, a = 4m/s
2
, s= 0.5.
? 0.5 = 2t + (1/2) 4 t
2
? 2 t
2
+ 2 t – 0.5 =0
(body moving us)
?R
R
???
F
mg
(body moving down)
?R
R
???
mg
R
45°
?R
mg
F
30°
?R
R
30°
mg
mg
ma
?R
R
Chapter 6
6.3
? 4 t
2
+ 4 t – 1 = 0
? = 
4 2
16 16 4
?
? ? ?
= 
8
656 . 1
= 0.207sec
Time taken to cover next half meter is 0.21sec.
10. f ? applied force
F
i
? contact force
F ? frictional force
R ? normal reaction
? = tan ? = F/R
When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction 
increase upto limiting friction ( ?R)
Before reaching limiting friction
F < ?R
? tan ? = 
R
R
R
F ?
? ? tan ? ? ? ? ??? tan
–1
? ?
11. From the free body diagram
T + 0.5a – 0.5 g = 0 ...(1)
?R + 1a + T
1
– T = 0 ...(2)
?R + 1a – T
1
= 0
?R + 1a = T
1
...(3)
From (2) & (3) ? ?R + a =  T – T
1
? T – T
1
= T
1
? T = 2T
1
Equation (2) becomes ?R + a + T
1
– 2T
1
= 0
? ?R + a – T
1
= 0
? T
1
= ?R + a = 0.2g + a ...(4)
Equation (1) becomes 2T
1
+ 0/5a – 0.5g = 0
? T
1
= 
2
a 5 . 0 g 5 . 0 ?
= 0.25g – 0.25a ...(5)
From (4) & (5)  0.2g + a = 0.25g – 0.25a 
? a = 
25 . 1
05 . 0
× 10 = 0.04 I 10 = 0.4m/s
2
a) Accln of 1kg blocks each is 0.4m/s
2
b) Tension T
1
= 0.2g + a + 0.4 = 2.4N
c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N
12. From the free body diagram
?
1
R + 1 – 16 = 0
? ?
1
(2g) + (–15) = 0
? ?
1
= 15/20 = 0.75
?
2
R
1
+ 4 × 0.5 + 16 – 4g sin 30° = 0
??
2
(20 3 ) + 2 + 16  – 20 = 0
? ?
2
= 
3 20
2
= 
32 . 17
1
= 0.057 ? 0.06
?Co-efficient of friction ?
1
= 0.75 & ?
2
= 0.06 ?
F
f
? ?
Fi
R
Limiting
Friction
?R
T 1
R
A  1g
1a
?
?R
R
1g
1a
?
0.5g
0.5g
a
B A
?=0.2
0.5kg
1kg 1kg
?=0.2
2×0.5
16N
?R 1
4g
16N=T
? 2R
4×0.5
30°
a
? 1
0.5 m/s
2
2kg
4kg
? 2
Chapter 6
6.4
13.
From the free body diagram
T + 15a – 15g = 0 T – (T
1
+ 5a+ ?R)= 0 T
1
– 5g – 5a = 0 ?
? T = 15g – 15 a   ...(i) ? T – (5g + 5a + 5a + ?R) = 0 ?T
1
=5g + 5a  …(iii)
? T = 5g + 10a + ?R  …(ii) ?
From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g)
? 25a = 90 ? a = 3.6m/s
2
Equation (ii) ? T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10
? 96N in the left string
Equation (iii) T
1
= 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string.
14. s = 5m, ? = 4/3, g = 10m/s
2
u = 36km/h = 10m/s, v = 0, 
a = 
s 2
u v
2 2
?
= 
5 2
10 0
2
?
?
= –10m/s
2
From the freebody diagrams,
R – mg cos ? = 0 ; g = 10m/s
2
? R = mg cos ??….(i) ; ? = 4/3.
Again, ma + mg sin ? - ? R = 0
? ma + mg sin ? – ? mg cos ? = 0
??a + g sin ? – mg cos ? = 0
? 10 + 10 sin ? - (4/3) × 10 cos ? = 0
? 30 + 30 sin ? – 40 cos ? =0
? 3 + 3 sin ? – 4 cos ? = 0
? 4 cos ? - 3 sin ? = 3
? 4 ? ?
2
sin 1 = 3 + 3 sin ?
? 16 (1 – sin
2 
?) = 9 + 9 sin
2 
??+ 18 sin ?
sin ? = 
25 2
) 7 )( 25 ( 4 18 18
2
?
? ? ? ?
= 
50
32 18 ? ?
=
50
14
= 0.28 [Taking +ve sign only]
? ? = sin
–1
(0.28) = 16°
Maximum incline is ? = 16° ?
15. to reach in minimum time, he has to move with maximum possible acceleration. 
Let, the maximum acceleration is  ‘a’
? ma – ?R = 0 ? ma = ? mg
? a = ? g = 0.9 × 10 = 9m/s
2
a) Initial velocity u = 0, t = ?
a = 9m/s
2
, s = 50m
s = ut + ½ at
2
? 50 = 0 + (1/2) 9 t
2
? t = 
9
100
= 
3
10
sec.
b) After overing 50m, velocity of the athelete is
V = u + at = 0 + 9 × (10/3) = 30m/s
He has to stop in minimum time. So deceleration ia –a = –9m/s
2
(max)
T 1
5g
5a
T 
15g
15a
A 
B
R
5g
T 1
T
r=5g
?R
C
B
A
a
15kg 15kg
a
a
R
ma
mg
?R
a
? ?
velocity
??? the max.
           angle
?R
a
R 
mg
ma
?R
a
R 
mg
ma
Page 5


6.1
SOLUTIONS TO CONCEPTS 
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0 ? R = mg ...(1)
Again ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 ?
2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0 ? R = mg ...(1)
ma – ? R = 0 ? ma = ? R = ? mg (from (1))
? a = ?g = 0.1 × 10 = 1m/s
2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s
2
(deceleration)
S = 
a 2
u v
2 2
?
= 
) 1 ( 2
10 0
2
?
?
= 
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f ? frictional force
F ? Applied force
From grap it can be seen that when applied force is zero, 
frictional force is zero.
4. From the free body diagram,
R – mg cos ? = 0 ? R = mg cos ? ..(1)
For the block
U = 0, s = 8m, t = 2sec.
?s = ut + ½ at
2
? 8 = 0 + ½ a 2
2
? a = 4m/s
2
Again, ?R + ma – mg sin ? = 0
? ? mg cos ? + ma – mg sin ? = 0 [from (1)]
? m( ?g cos ? + a – g sin ?) = 0
? ? × 10 × cos 30° = g sin 30° – a
??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4
? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11
? Co-efficient of kinetic friction between the two is 0.11.
5. From the free body diagram
4 – 4a – ?R + 4g sin 30° = 0 …(1) 
R – 4g cos 30° = 0 ...(2)
? R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0
? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s
2
For the block u =0, t = 2sec, a = 5m/s
2
Distance s = ut + ½ at
2 
? s = 0 + (1/2) 5 × 2
2
= 10m
The block will move 10m.
R
mg
ma
?R
a
R
mg
ma
?R
a velocity
F
p
o
mg
R
30°
R
???
?R
mg
ma
4kg
30°
4N
R
???
?R
mg
ma
a velocity
Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to 
the net force acting down the incline = ? R + 2 g sin 30°
= 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net 
force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – ?R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
? Force required is zero.
7. From the free body diagram
g = 10m/s
2
, m = 2kg, ? = 30°, ? = 0.2
R – mg cos ? - F sin ? = 0
? R = mg cos ? + F sin ?? ...(1)
And mg sin ? + ?R – F cos ? = 0
? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0
? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0
? F =
) cos sin (
) cos mg sin mg (
? ? ? ?
? ? ? ?
? F = 
) 2 / 3 ( ) 2 / 1 ( 2 . 0
) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2
? ?
? ? ? ? ? ?
= 
76 . 0
464 . 13
= 17.7N ? 17.5N ?
8. m ? mass of child
R – mg cos 45° = 0
? R = mg cos 45° = mg /v
2
...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - ?R
= mg sin 45° - ? mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration  = 
mass
Force
= 
m
) 2 2 ( m
= 2 2 m/s
2
?
9. Suppose, the body is accelerating down with acceleration ‘a’.
From the free body diagram
R – mg cos ? = 0
? R = mg cos ? ...(1)
ma + mg sin ? – ? R = 0
? a = 
m
) cos (sin mg ? ? ? ?
= g (sin ? – ? cos ?)
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at
2
? 0.5 = 0 + ½ a (0/5)
2
? a = 4m/s
2
...(2)
For the next half metre 
u` = 2m/s, a = 4m/s
2
, s= 0.5.
? 0.5 = 2t + (1/2) 4 t
2
? 2 t
2
+ 2 t – 0.5 =0
(body moving us)
?R
R
???
F
mg
(body moving down)
?R
R
???
mg
R
45°
?R
mg
F
30°
?R
R
30°
mg
mg
ma
?R
R
Chapter 6
6.3
? 4 t
2
+ 4 t – 1 = 0
? = 
4 2
16 16 4
?
? ? ?
= 
8
656 . 1
= 0.207sec
Time taken to cover next half meter is 0.21sec.
10. f ? applied force
F
i
? contact force
F ? frictional force
R ? normal reaction
? = tan ? = F/R
When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction 
increase upto limiting friction ( ?R)
Before reaching limiting friction
F < ?R
? tan ? = 
R
R
R
F ?
? ? tan ? ? ? ? ??? tan
–1
? ?
11. From the free body diagram
T + 0.5a – 0.5 g = 0 ...(1)
?R + 1a + T
1
– T = 0 ...(2)
?R + 1a – T
1
= 0
?R + 1a = T
1
...(3)
From (2) & (3) ? ?R + a =  T – T
1
? T – T
1
= T
1
? T = 2T
1
Equation (2) becomes ?R + a + T
1
– 2T
1
= 0
? ?R + a – T
1
= 0
? T
1
= ?R + a = 0.2g + a ...(4)
Equation (1) becomes 2T
1
+ 0/5a – 0.5g = 0
? T
1
= 
2
a 5 . 0 g 5 . 0 ?
= 0.25g – 0.25a ...(5)
From (4) & (5)  0.2g + a = 0.25g – 0.25a 
? a = 
25 . 1
05 . 0
× 10 = 0.04 I 10 = 0.4m/s
2
a) Accln of 1kg blocks each is 0.4m/s
2
b) Tension T
1
= 0.2g + a + 0.4 = 2.4N
c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N
12. From the free body diagram
?
1
R + 1 – 16 = 0
? ?
1
(2g) + (–15) = 0
? ?
1
= 15/20 = 0.75
?
2
R
1
+ 4 × 0.5 + 16 – 4g sin 30° = 0
??
2
(20 3 ) + 2 + 16  – 20 = 0
? ?
2
= 
3 20
2
= 
32 . 17
1
= 0.057 ? 0.06
?Co-efficient of friction ?
1
= 0.75 & ?
2
= 0.06 ?
F
f
? ?
Fi
R
Limiting
Friction
?R
T 1
R
A  1g
1a
?
?R
R
1g
1a
?
0.5g
0.5g
a
B A
?=0.2
0.5kg
1kg 1kg
?=0.2
2×0.5
16N
?R 1
4g
16N=T
? 2R
4×0.5
30°
a
? 1
0.5 m/s
2
2kg
4kg
? 2
Chapter 6
6.4
13.
From the free body diagram
T + 15a – 15g = 0 T – (T
1
+ 5a+ ?R)= 0 T
1
– 5g – 5a = 0 ?
? T = 15g – 15 a   ...(i) ? T – (5g + 5a + 5a + ?R) = 0 ?T
1
=5g + 5a  …(iii)
? T = 5g + 10a + ?R  …(ii) ?
From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g)
? 25a = 90 ? a = 3.6m/s
2
Equation (ii) ? T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10
? 96N in the left string
Equation (iii) T
1
= 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string.
14. s = 5m, ? = 4/3, g = 10m/s
2
u = 36km/h = 10m/s, v = 0, 
a = 
s 2
u v
2 2
?
= 
5 2
10 0
2
?
?
= –10m/s
2
From the freebody diagrams,
R – mg cos ? = 0 ; g = 10m/s
2
? R = mg cos ??….(i) ; ? = 4/3.
Again, ma + mg sin ? - ? R = 0
? ma + mg sin ? – ? mg cos ? = 0
??a + g sin ? – mg cos ? = 0
? 10 + 10 sin ? - (4/3) × 10 cos ? = 0
? 30 + 30 sin ? – 40 cos ? =0
? 3 + 3 sin ? – 4 cos ? = 0
? 4 cos ? - 3 sin ? = 3
? 4 ? ?
2
sin 1 = 3 + 3 sin ?
? 16 (1 – sin
2 
?) = 9 + 9 sin
2 
??+ 18 sin ?
sin ? = 
25 2
) 7 )( 25 ( 4 18 18
2
?
? ? ? ?
= 
50
32 18 ? ?
=
50
14
= 0.28 [Taking +ve sign only]
? ? = sin
–1
(0.28) = 16°
Maximum incline is ? = 16° ?
15. to reach in minimum time, he has to move with maximum possible acceleration. 
Let, the maximum acceleration is  ‘a’
? ma – ?R = 0 ? ma = ? mg
? a = ? g = 0.9 × 10 = 9m/s
2
a) Initial velocity u = 0, t = ?
a = 9m/s
2
, s = 50m
s = ut + ½ at
2
? 50 = 0 + (1/2) 9 t
2
? t = 
9
100
= 
3
10
sec.
b) After overing 50m, velocity of the athelete is
V = u + at = 0 + 9 × (10/3) = 30m/s
He has to stop in minimum time. So deceleration ia –a = –9m/s
2
(max)
T 1
5g
5a
T 
15g
15a
A 
B
R
5g
T 1
T
r=5g
?R
C
B
A
a
15kg 15kg
a
a
R
ma
mg
?R
a
? ?
velocity
??? the max.
           angle
?R
a
R 
mg
ma
?R
a
R 
mg
ma
Chapter 6
6.5
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ?
?
) on Decelerati ( s / m 9 g a
) force frictional (max R ma
ma R
2
?
u
1
= 30m/s, v
1
= 0
t = 
a
u v
1 1
?
= 
a
30 0
?
?
= 
a
30
?
?
= 
3
10
sec.
16. Hardest brake means maximum force of friction is developed between car’s type & road.
Max frictional force = ?R
From the free body diagram
R – mg cos ? =0
? R = mg cos ? ...(i)
and ?R + ma – mg sin ) = 0 …(ii)
? ?mg cos ? + ma – mg sin ? = 0
? ?g cos ? + a – 10 × (1/2) = 0
??a = 5 – {1 – (2 3 )} × 10 ( 2 / 3 ) = 2.5 m/s
2
When, hardest brake is applied the car move with acceleration 2.5m/s
2
S = 12.8m, u = 6m/s
S0, velocity at the end of incline
V = as 2 u
2
? = ) 8 . 12 )( 5 . 2 ( 2 6
2
? = 64 36 ? = 10m/s = 36km/h
Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36km/h.
17. Let, , a maximum acceleration produced in car.
? ma = ?R [For more acceleration, the tyres will slip]
? ma = ? mg ? a = ?g = 1 × 10 = 10m/s
2
For crossing the bridge in minimum time, it has to travel with maximum 
acceleration
u = 0, s = 500m, a = 10m/s
2
s = ut + ½ at
2
? 500 = 0 + (1/2) 10 t
2
? t = 10 sec.
If acceleration is less than 10m/s
2
, time will be more than 10sec. So one can’t drive through the bridge 
in less than 10sec. ?
18. From the free body diagram
R = 4g cos 30° = 4 × 10  × 2 / 3 = 20 3 ...(i)
?
2
R + 4a – P – 4g sin 30° = 0 ? 0.3 (40) cos 30°  + 4a – P – 40 sin 20° = 0 …(ii)
P + 2a + ?
1
R
1
– 2g sin 30° = 0 …(iii)
R
1
= 2g cos 30° = 2 × 10 × 2 / 3 = 10 3 ...(iv)
Equn. (ii) 6 3 + 4a – P – 20 = 0
Equn (iv) P + 2a + 2 3 – 10 = 0
From Equn (ii) & (iv) 6 3 + 6a – 30 + 2 3 = 0
? 6a = 30 – 8 3 = 30 – 13.85 = 16.15
? a = 
6
15 . 16
= 2.69 = 2.7m/s
2
b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of 
friction is less than that of 2kg. So, they will move separately. Drawing the free body diagram of 2kg 
mass only, it can be found that, a = 2.4m/s
2
. ?
ma
R
a
mg
?R
?R
a
mg
R
4kg
30°
2kg
a
?a
R
4g
? ?R
P
2a
R
2g
P
? 1 R 1 ?