# H C Verma Solutions

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 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/s2, 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 =

( sin cos )

(mgsin mgcos )

F =

0.2 (1/ 2) ( 3 / 2)

2 10 (1/ 2) 0.2 2 10 ( 3 / 2)

=

0.76

13.464

= 17.7N 17.5N

8. m mass of child

R – mg cos 45° = 0

R = mg cos 45° = mg /v2 ...(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

m(2 2)

= 2 2 m/s2

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

mg(sin cos )

= 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 at2 0.5 = 0 + ½ a (0/5)2 a = 4m/s2 ...(2) For the next half metre

u` = 2m/s, a = 4m/s2, s= 0.5.

0.5 = 2t + (1/2) 4 t2 2 t2 + 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 t2 + 4 t – 1 = 0

=

2 4

4 16 16

=

8

1.656

= 0.207sec

Time taken to cover next half meter is 0.21sec.

10. f applied force

Fi contact force

F frictional force

R normal...

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