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__Introductory Exercise 5.2__

**Ques: 1 Three blocks of mass 1 kg, 4 kg and 2 kg are placed on a smooth horizontal plane as shown in figure. Find: ****(a) the acceleration of the system, ****(b) the normal force between 1 kg block and 4 kg block, ****(c) the net force on 2 kg block.****Ans: **(a) 10 ms^{-2} (b) 110 N (c) 20 N**Sol:** Acceleration of system

= 10 m/s^{2}

Let normal force between 1 kg block and 4 kg block = F_{1}

âˆ´ Net force on 1 kg block = 120 - N

âˆ´

or 10 = 120 - F_{1}

i.e., F_{1} = 110 N

Net force on 2 kg block = 2*a

= 2*10

=20 N**Ques 2: Two blocks of mass 2 kg and 4 kg are released from rest over a smooth inclined plane of inclination 30Â° as shown in figure. What is the normal force between the two blocks?Ans:** zero

The normal force between the two blocks will be zero.

Ans:

âˆ´ 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

â‡’

Ans:

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

Ans:

T cos Ï† = mg + ma cos (90Â° - Î¸)

i.e., T cos Ï† = mg + ma sin Î¸ â€¦(i)

and T sin Ï† = ma cos Î¸

Squaring and adding Eqs. (i) and (ii),

T

T

or

Dividing Eq. (i) by Eq. (ii),

i.e.,

Ans:

Net force on 1 kg mass = 8 - T

âˆ´ 8 - T

â‡’ T

Net force on 1 kg block = T

âˆ´ T

__Introductory Exercise 5.3__

**Ques 7: In the arrangement shown in figure what should be the mass of block A, so that the system remains at rest? Neglect friction and mass of strings.Ans:** 3 kg

For the system to remain at rest

T

T

or T

T

Substituting the values of T

2 g + g = mg

i.e., m = 3 kg

Ans:

Sol:

âˆ´ T

and T

âˆ´

Ans:

T - 1 g = 1a

Adding above two equations

1 g = 3 a

âˆ´ a = g/3

Velocity of 1kg block 1 section after the system is set in motion

v = 0 + at

On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t' ) taken by the 1 k g block to attain zero velocity will be given by the equation.

â‡’

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 time upward displacement of 1 kg block

Downward displacement of 2 kg block

As the two are just equal, the string will again become taut after time 1/3 s.

Ans:

and T - 2g = 2a â€¦(ii)

Adding Eqs. (i) and (ii),

F - 1g = 3a

âˆ´

__Introductory Exercise 5.2__

**Ques 11: Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are smooth.(a) Find the acceleration of 1 kg block.(b) Find the tension in the string.(g = 10 m/s ** (a) 2g/3, (b) 10/3 N

and 1g - T = 2a â€¦(ii)

Solving Eqs. (i) and (ii),

Î± = g/3

âˆ´ Acceleration of 1 kg block

Tension in the string

Ans:

T = Ma â€¦(ii)

Solving Eqs. (i) and (ii)

Î± = g/2

and T = Mg

Ans:

T = Mg â€¦(i)

For the motion of block of mass 3 kg

....(ii)

For the motion of block of mass 2 kg

....(iii)

Adding Eqs. (ii) and (iii),

g = 5a

i.e., Î± = g/5

Substituting above value of a in Eq. (iii),

T/2 = 2(g+a)

or

= 24g/5

Substituting value of above value of T in Eq. (i),

M = 24/5

= 4.8 kg

Ans:

i.e., T = 4 m

F - T = m

or F - 4m

or

âˆ´ T = 4 m

= 4*0.3* 2/7

= 2.4/7

= 12/35 N

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