H C Verma Solutions

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/s2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s a = –1m/s2 (deceleration)
S =
2a
v2  u2
=
2( 1)
0 102


=
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 at2  8 = 0 + ½ a 22  a = 4m/s2
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
 (5 / 3)  =1   = 1/ (5 / 3) = 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 × ( 3 / 2) + 4 × 10 × (1/2) = 0
 4 – 4a – 3.81 + 20 = 0  a  5 m/s2
For the block u =0, t = 2sec, a = 5m/s2
Distance s = ut + ½ at2  s = 0 + (1/2) 5 × 22 = 10m
The block will move 10m.
R
mg ma R a R mg ma R velocity a
F
p o mg
R
30°
R

R
mg ma 4kg
30°
4N
R

R
mg ma velocity a
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