QBManager

Laws of Motion

146035 Two block $A$ and $B$ of masses $1.5 kg$ and $0.5 kg$ , respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block $A$ is lifted until block B touches the ground and then block $A$ is released. The initial height of block $A$ is $80 cm$ when block $B$ just touches the ground. The maximum height reached by block $B$ from the ground after the block $A$ falls on the ground is

1

80 cm

2

120 cm

3

140 cm

4

160 cm

Explanation:

B
From F.B.D. of the system, the common acceleration of two masses,
$a = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) g$
$= (\frac{1.5 - 0.5}{1.5 + 0.5}) 10$
$a = 5 m / s^{2}$
When block A is released,
$v^{2} = u^{2} + 2 gh$
$v^{2} = 0 + 2 \times 10 h$
$v^{2} = 20 h$ .
But for height $0.8 m$ , block A falls with common acceleration, therefore from third ${eq}^{n}$ . of motion,
$v_{1}^{2} = u_{1}^{2} - 2 a \times 0.8$
Hence, $v_{1} = 0, u_{1}^{2} = v^{2} = 20 h$
$0 = 20 \times h - 2 \times 5 \times 0.8 (∵ a = 5 m / s^{2})$
$20 h = 8$
$h = 0.4 m = 40 cm .$
Therefore, maximum height reached by block B
$= 80 + 40 = 120 cm$

AP EAMCET (21.04.2019) Shift-II

Laws of Motion

146036 Two blocks of masses ' $m_{1}$ ' and ' $m_{2}$ ' are suspended by a massless string passing over a smooth pulley, if the acceleration of the system
is $\frac{g}{8}$ then the ratio of the masses $\frac{m_{2}}{m_{1}} =$

1

8 : 1

2

8 : 7

3

9 : 7

4

4 : 3

Explanation:

C Given, $a = g / 8$

FBD for block $m_{1}$ ,

$T = m_{1} a + m_{1} g$
FBD for block $m_{2}$ ,

$T = m_{2} g - m_{2} a$
From equation (i) $i i$ , we get-
$m_{2} g - m_{1} g = m_{2} a + m_{1} a$
$g (m_{2} - m_{1}) = a (m_{2} + m_{1})$
$a = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{1}{8} = \frac{m_{2} - m_{1}}{m_{2} + m_{1}}$
$m_{2} + m_{1} = 8 m_{2} - 8 m_{1}$
$9 m_{1} = 7 m_{2}$
$\frac{m_{2}}{m_{1}} = \frac{9}{7}$

AP EAMCET-25.04.2018

Laws of Motion

146037 A man weighing $60 kg$ is in lift moving down with an acceleration of $1.8 {ms}^{- 2}$ . The force exerted by the floor on him is :

1

588 N

2

480 N

3 zero

4

696 N

Explanation:

B
Given that,
Acceleration (a) $= 1.8 m / s^{2}$
mass $(m) = 60 kg$
Now,
$N + F_{P} = mg$
$N + ma = mg (∵ F_{P} = m \times a)$
Then,
$Normal (N) = mg - ma$
$= m (g - a)$
$= 60 \times (9.8 - 1.8)$
$= 480 N$

Karnataka CET-2018

Laws of Motion

146038 If the block moves up with constant velocity $v$ $m / s$ , then find $F$ .

1

F = \frac{mg}{2}

2

F = \frac{2 m g}{3}

3

F = \frac{mg}{3}

4

F = \frac{5 mg}{3}

Explanation:

C Since the block is moving with constant velocity thus, the net force acting on the block will be zero.
So, the tension in the wire will be balanced by weight
$3 F = mg$
$F = mg / 3$

AIIMS-26.05.2018(E)

Laws of Motion

146035 Two block $A$ and $B$ of masses $1.5 kg$ and $0.5 kg$ , respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block $A$ is lifted until block B touches the ground and then block $A$ is released. The initial height of block $A$ is $80 cm$ when block $B$ just touches the ground. The maximum height reached by block $B$ from the ground after the block $A$ falls on the ground is

1

80 cm

2

120 cm

3

140 cm

4

160 cm

Explanation:

B
From F.B.D. of the system, the common acceleration of two masses,
$a = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) g$
$= (\frac{1.5 - 0.5}{1.5 + 0.5}) 10$
$a = 5 m / s^{2}$
When block A is released,
$v^{2} = u^{2} + 2 gh$
$v^{2} = 0 + 2 \times 10 h$
$v^{2} = 20 h$ .
But for height $0.8 m$ , block A falls with common acceleration, therefore from third ${eq}^{n}$ . of motion,
$v_{1}^{2} = u_{1}^{2} - 2 a \times 0.8$
Hence, $v_{1} = 0, u_{1}^{2} = v^{2} = 20 h$
$0 = 20 \times h - 2 \times 5 \times 0.8 (∵ a = 5 m / s^{2})$
$20 h = 8$
$h = 0.4 m = 40 cm .$
Therefore, maximum height reached by block B
$= 80 + 40 = 120 cm$

AP EAMCET (21.04.2019) Shift-II

Laws of Motion

146036 Two blocks of masses ' $m_{1}$ ' and ' $m_{2}$ ' are suspended by a massless string passing over a smooth pulley, if the acceleration of the system
is $\frac{g}{8}$ then the ratio of the masses $\frac{m_{2}}{m_{1}} =$

1

8 : 1

2

8 : 7

3

9 : 7

4

4 : 3

Explanation:

C Given, $a = g / 8$

FBD for block $m_{1}$ ,

$T = m_{1} a + m_{1} g$
FBD for block $m_{2}$ ,

$T = m_{2} g - m_{2} a$
From equation (i) $i i$ , we get-
$m_{2} g - m_{1} g = m_{2} a + m_{1} a$
$g (m_{2} - m_{1}) = a (m_{2} + m_{1})$
$a = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{1}{8} = \frac{m_{2} - m_{1}}{m_{2} + m_{1}}$
$m_{2} + m_{1} = 8 m_{2} - 8 m_{1}$
$9 m_{1} = 7 m_{2}$
$\frac{m_{2}}{m_{1}} = \frac{9}{7}$

AP EAMCET-25.04.2018

Laws of Motion

146037 A man weighing $60 kg$ is in lift moving down with an acceleration of $1.8 {ms}^{- 2}$ . The force exerted by the floor on him is :

1

588 N

2

480 N

3 zero

4

696 N

Explanation:

B
Given that,
Acceleration (a) $= 1.8 m / s^{2}$
mass $(m) = 60 kg$
Now,
$N + F_{P} = mg$
$N + ma = mg (∵ F_{P} = m \times a)$
Then,
$Normal (N) = mg - ma$
$= m (g - a)$
$= 60 \times (9.8 - 1.8)$
$= 480 N$

Karnataka CET-2018

Laws of Motion

146038 If the block moves up with constant velocity $v$ $m / s$ , then find $F$ .

1

F = \frac{mg}{2}

2

F = \frac{2 m g}{3}

3

F = \frac{mg}{3}

4

F = \frac{5 mg}{3}

Explanation:

C Since the block is moving with constant velocity thus, the net force acting on the block will be zero.
So, the tension in the wire will be balanced by weight
$3 F = mg$
$F = mg / 3$

AIIMS-26.05.2018(E)

Laws of Motion

146035 Two block $A$ and $B$ of masses $1.5 kg$ and $0.5 kg$ , respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block $A$ is lifted until block B touches the ground and then block $A$ is released. The initial height of block $A$ is $80 cm$ when block $B$ just touches the ground. The maximum height reached by block $B$ from the ground after the block $A$ falls on the ground is

1

80 cm

2

120 cm

3

140 cm

4

160 cm

Explanation:

B
From F.B.D. of the system, the common acceleration of two masses,
$a = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) g$
$= (\frac{1.5 - 0.5}{1.5 + 0.5}) 10$
$a = 5 m / s^{2}$
When block A is released,
$v^{2} = u^{2} + 2 gh$
$v^{2} = 0 + 2 \times 10 h$
$v^{2} = 20 h$ .
But for height $0.8 m$ , block A falls with common acceleration, therefore from third ${eq}^{n}$ . of motion,
$v_{1}^{2} = u_{1}^{2} - 2 a \times 0.8$
Hence, $v_{1} = 0, u_{1}^{2} = v^{2} = 20 h$
$0 = 20 \times h - 2 \times 5 \times 0.8 (∵ a = 5 m / s^{2})$
$20 h = 8$
$h = 0.4 m = 40 cm .$
Therefore, maximum height reached by block B
$= 80 + 40 = 120 cm$

AP EAMCET (21.04.2019) Shift-II

Laws of Motion

146036 Two blocks of masses ' $m_{1}$ ' and ' $m_{2}$ ' are suspended by a massless string passing over a smooth pulley, if the acceleration of the system
is $\frac{g}{8}$ then the ratio of the masses $\frac{m_{2}}{m_{1}} =$

1

8 : 1

2

8 : 7

3

9 : 7

4

4 : 3

Explanation:

C Given, $a = g / 8$

FBD for block $m_{1}$ ,

$T = m_{1} a + m_{1} g$
FBD for block $m_{2}$ ,

$T = m_{2} g - m_{2} a$
From equation (i) $i i$ , we get-
$m_{2} g - m_{1} g = m_{2} a + m_{1} a$
$g (m_{2} - m_{1}) = a (m_{2} + m_{1})$
$a = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{1}{8} = \frac{m_{2} - m_{1}}{m_{2} + m_{1}}$
$m_{2} + m_{1} = 8 m_{2} - 8 m_{1}$
$9 m_{1} = 7 m_{2}$
$\frac{m_{2}}{m_{1}} = \frac{9}{7}$

AP EAMCET-25.04.2018

Laws of Motion

146037 A man weighing $60 kg$ is in lift moving down with an acceleration of $1.8 {ms}^{- 2}$ . The force exerted by the floor on him is :

1

588 N

2

480 N

3 zero

4

696 N

Explanation:

B
Given that,
Acceleration (a) $= 1.8 m / s^{2}$
mass $(m) = 60 kg$
Now,
$N + F_{P} = mg$
$N + ma = mg (∵ F_{P} = m \times a)$
Then,
$Normal (N) = mg - ma$
$= m (g - a)$
$= 60 \times (9.8 - 1.8)$
$= 480 N$

Karnataka CET-2018

Laws of Motion

146038 If the block moves up with constant velocity $v$ $m / s$ , then find $F$ .

1

F = \frac{mg}{2}

2

F = \frac{2 m g}{3}

3

F = \frac{mg}{3}

4

F = \frac{5 mg}{3}

Explanation:

C Since the block is moving with constant velocity thus, the net force acting on the block will be zero.
So, the tension in the wire will be balanced by weight
$3 F = mg$
$F = mg / 3$

AIIMS-26.05.2018(E)

Laws of Motion

146035 Two block $A$ and $B$ of masses $1.5 kg$ and $0.5 kg$ , respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block $A$ is lifted until block B touches the ground and then block $A$ is released. The initial height of block $A$ is $80 cm$ when block $B$ just touches the ground. The maximum height reached by block $B$ from the ground after the block $A$ falls on the ground is

1

80 cm

2

120 cm

3

140 cm

4

160 cm

Explanation:

B
From F.B.D. of the system, the common acceleration of two masses,
$a = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) g$
$= (\frac{1.5 - 0.5}{1.5 + 0.5}) 10$
$a = 5 m / s^{2}$
When block A is released,
$v^{2} = u^{2} + 2 gh$
$v^{2} = 0 + 2 \times 10 h$
$v^{2} = 20 h$ .
But for height $0.8 m$ , block A falls with common acceleration, therefore from third ${eq}^{n}$ . of motion,
$v_{1}^{2} = u_{1}^{2} - 2 a \times 0.8$
Hence, $v_{1} = 0, u_{1}^{2} = v^{2} = 20 h$
$0 = 20 \times h - 2 \times 5 \times 0.8 (∵ a = 5 m / s^{2})$
$20 h = 8$
$h = 0.4 m = 40 cm .$
Therefore, maximum height reached by block B
$= 80 + 40 = 120 cm$

AP EAMCET (21.04.2019) Shift-II

Laws of Motion

146036 Two blocks of masses ' $m_{1}$ ' and ' $m_{2}$ ' are suspended by a massless string passing over a smooth pulley, if the acceleration of the system
is $\frac{g}{8}$ then the ratio of the masses $\frac{m_{2}}{m_{1}} =$

1

8 : 1

2

8 : 7

3

9 : 7

4

4 : 3

Explanation:

C Given, $a = g / 8$

FBD for block $m_{1}$ ,

$T = m_{1} a + m_{1} g$
FBD for block $m_{2}$ ,

$T = m_{2} g - m_{2} a$
From equation (i) $i i$ , we get-
$m_{2} g - m_{1} g = m_{2} a + m_{1} a$
$g (m_{2} - m_{1}) = a (m_{2} + m_{1})$
$a = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{2} + m_{1}}$
$\frac{1}{8} = \frac{m_{2} - m_{1}}{m_{2} + m_{1}}$
$m_{2} + m_{1} = 8 m_{2} - 8 m_{1}$
$9 m_{1} = 7 m_{2}$
$\frac{m_{2}}{m_{1}} = \frac{9}{7}$

AP EAMCET-25.04.2018

Laws of Motion

146037 A man weighing $60 kg$ is in lift moving down with an acceleration of $1.8 {ms}^{- 2}$ . The force exerted by the floor on him is :

1

588 N

2

480 N

3 zero

4

696 N

Explanation:

B
Given that,
Acceleration (a) $= 1.8 m / s^{2}$
mass $(m) = 60 kg$
Now,
$N + F_{P} = mg$
$N + ma = mg (∵ F_{P} = m \times a)$
Then,
$Normal (N) = mg - ma$
$= m (g - a)$
$= 60 \times (9.8 - 1.8)$
$= 480 N$

Karnataka CET-2018

Laws of Motion

146038 If the block moves up with constant velocity $v$ $m / s$ , then find $F$ .

1

F = \frac{mg}{2}

2

F = \frac{2 m g}{3}

3

F = \frac{mg}{3}

4

F = \frac{5 mg}{3}

Explanation:

C Since the block is moving with constant velocity thus, the net force acting on the block will be zero.
So, the tension in the wire will be balanced by weight
$3 F = mg$
$F = mg / 3$

AIIMS-26.05.2018(E)

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