QBManager

LAWS OF MOTION (ADDITIONAL)

372240 Two blocks of masses $m$ and $2 m$ are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass $m$ is placed on a smooth inclined plane of inclination $30^{\circ}$ and $2 m$ hangs vertically. If the system is released, the blocks move with an acceleration equal to

1

g / 4

2

g / 3

3

g / 2

4

g

Explanation:

C In downward,
$2 mg - T = 2 ma$
And
$mg \sin 30^{\circ} - T = - ma$

Subtracting both equation,
$2 mg - \frac{mg}{2} = 3 ma$
$3 a = \frac{ug - g}{2}$
$3 a = \frac{3 g}{2}$
$a = \frac{g}{2}$

AMU-2009

LAWS OF MOTION (ADDITIONAL)

372241 In the given figure, three blocks are connected by massless cords. Masses of the block are $m$ , $3 m$ ad $5 m$ and they are pulled by a force $F$ on a frictionless horizontal surface. If the tension $T_{1} = 16 N$ , then $T_{2}$ is equal to :

1

6 N

2

10 N

3

(80 / 3) N

4

18 N

Explanation:

B Given, $T_{1} = 16 N$ ,
$m_{1} = m, m_{2} = 3 m, m_{3} = 5 m$
$T_{2} = ?$
Acceleration " $a$ " due to tension $T_{1}$ in mass $m_{2}$ and $m_{3}$
$a = \frac{T_{1}}{m_{2} + m_{3}} = \frac{16}{3 m + 5 m}$
$a = \frac{2}{m}$
Now, tension
$T_{2} = m_{3} \times a$
$T_{2} = 5 m \times \frac{2}{m}$
$T_{2} = 10 N$
So, tension $T_{2}$ is equal to $10 N$ .

AMU-2006

LAWS OF MOTION (ADDITIONAL)

372242 The apparent weight of a person in a lift moving downwards is half his apparent weight the same lift moving upwards with the same acceleration. Acceleration of the lift is

1

g

2

\frac{g}{4}

3

\frac{g}{2}

4

\frac{g}{3}

Explanation:

D Let's,
Mass of person $= m$
Acceleration $= a$
During downward and upward moving
Then,
Apparent weight during downward moving $(W_{down})$
$W_{down} = m (g - a)$
Apparent weight during upwards moving $(W_{up})$
$W_{up} = m (g + a)$
Given,
$W_{down} = \frac{1}{2} W_{up}$
Then, $m (g - a) = \frac{1}{2} m (g + a)$
$a = \frac{g}{3}$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372243 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only

2 B only

3 Both A and B

4 neither A nor B

Explanation:

A All masses are pushed by force $F$

Then, acceleration-
$a = \frac{F}{total mass} = \frac{F}{m_{1} + m_{2} + m_{3}} = \frac{F}{3 + 2 + 1}$
$a = \frac{F}{6} m / s^{2}$
So, All block move with same acceleration $\frac{F}{6} m / s^{2}$
Net force on mass, $m_{1} = 3 kg$
$F_{1} = m_{1} \cdot a = 3 \cdot \frac{F}{6} = \frac{F}{2}$
$F_{1} = \frac{F}{2} N$
Net force on mass, $m_{2} = 2 kg$
$F_{2} = m_{2} \cdot a = 2 \cdot \frac{F}{6} = \frac{F}{3}$
$F_{2} = \frac{F}{3} N$
Net force on mass, $m_{3} = 1 kg$
$F_{3} = m_{3} \cdot a = 1 \times \frac{F}{6} = \frac{F}{6}$
$F_{3} = \frac{F}{6} N$
So, the net force on each block are different.

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372240 Two blocks of masses $m$ and $2 m$ are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass $m$ is placed on a smooth inclined plane of inclination $30^{\circ}$ and $2 m$ hangs vertically. If the system is released, the blocks move with an acceleration equal to

1

g / 4

2

g / 3

3

g / 2

4

g

Explanation:

C In downward,
$2 mg - T = 2 ma$
And
$mg \sin 30^{\circ} - T = - ma$

Subtracting both equation,
$2 mg - \frac{mg}{2} = 3 ma$
$3 a = \frac{ug - g}{2}$
$3 a = \frac{3 g}{2}$
$a = \frac{g}{2}$

AMU-2009

LAWS OF MOTION (ADDITIONAL)

372241 In the given figure, three blocks are connected by massless cords. Masses of the block are $m$ , $3 m$ ad $5 m$ and they are pulled by a force $F$ on a frictionless horizontal surface. If the tension $T_{1} = 16 N$ , then $T_{2}$ is equal to :

1

6 N

2

10 N

3

(80 / 3) N

4

18 N

Explanation:

B Given, $T_{1} = 16 N$ ,
$m_{1} = m, m_{2} = 3 m, m_{3} = 5 m$
$T_{2} = ?$
Acceleration " $a$ " due to tension $T_{1}$ in mass $m_{2}$ and $m_{3}$
$a = \frac{T_{1}}{m_{2} + m_{3}} = \frac{16}{3 m + 5 m}$
$a = \frac{2}{m}$
Now, tension
$T_{2} = m_{3} \times a$
$T_{2} = 5 m \times \frac{2}{m}$
$T_{2} = 10 N$
So, tension $T_{2}$ is equal to $10 N$ .

AMU-2006

LAWS OF MOTION (ADDITIONAL)

372242 The apparent weight of a person in a lift moving downwards is half his apparent weight the same lift moving upwards with the same acceleration. Acceleration of the lift is

1

g

2

\frac{g}{4}

3

\frac{g}{2}

4

\frac{g}{3}

Explanation:

D Let's,
Mass of person $= m$
Acceleration $= a$
During downward and upward moving
Then,
Apparent weight during downward moving $(W_{down})$
$W_{down} = m (g - a)$
Apparent weight during upwards moving $(W_{up})$
$W_{up} = m (g + a)$
Given,
$W_{down} = \frac{1}{2} W_{up}$
Then, $m (g - a) = \frac{1}{2} m (g + a)$
$a = \frac{g}{3}$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372243 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only

2 B only

3 Both A and B

4 neither A nor B

Explanation:

A All masses are pushed by force $F$

Then, acceleration-
$a = \frac{F}{total mass} = \frac{F}{m_{1} + m_{2} + m_{3}} = \frac{F}{3 + 2 + 1}$
$a = \frac{F}{6} m / s^{2}$
So, All block move with same acceleration $\frac{F}{6} m / s^{2}$
Net force on mass, $m_{1} = 3 kg$
$F_{1} = m_{1} \cdot a = 3 \cdot \frac{F}{6} = \frac{F}{2}$
$F_{1} = \frac{F}{2} N$
Net force on mass, $m_{2} = 2 kg$
$F_{2} = m_{2} \cdot a = 2 \cdot \frac{F}{6} = \frac{F}{3}$
$F_{2} = \frac{F}{3} N$
Net force on mass, $m_{3} = 1 kg$
$F_{3} = m_{3} \cdot a = 1 \times \frac{F}{6} = \frac{F}{6}$
$F_{3} = \frac{F}{6} N$
So, the net force on each block are different.

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372240 Two blocks of masses $m$ and $2 m$ are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass $m$ is placed on a smooth inclined plane of inclination $30^{\circ}$ and $2 m$ hangs vertically. If the system is released, the blocks move with an acceleration equal to

1

g / 4

2

g / 3

3

g / 2

4

g

Explanation:

C In downward,
$2 mg - T = 2 ma$
And
$mg \sin 30^{\circ} - T = - ma$

Subtracting both equation,
$2 mg - \frac{mg}{2} = 3 ma$
$3 a = \frac{ug - g}{2}$
$3 a = \frac{3 g}{2}$
$a = \frac{g}{2}$

AMU-2009

LAWS OF MOTION (ADDITIONAL)

372241 In the given figure, three blocks are connected by massless cords. Masses of the block are $m$ , $3 m$ ad $5 m$ and they are pulled by a force $F$ on a frictionless horizontal surface. If the tension $T_{1} = 16 N$ , then $T_{2}$ is equal to :

1

6 N

2

10 N

3

(80 / 3) N

4

18 N

Explanation:

B Given, $T_{1} = 16 N$ ,
$m_{1} = m, m_{2} = 3 m, m_{3} = 5 m$
$T_{2} = ?$
Acceleration " $a$ " due to tension $T_{1}$ in mass $m_{2}$ and $m_{3}$
$a = \frac{T_{1}}{m_{2} + m_{3}} = \frac{16}{3 m + 5 m}$
$a = \frac{2}{m}$
Now, tension
$T_{2} = m_{3} \times a$
$T_{2} = 5 m \times \frac{2}{m}$
$T_{2} = 10 N$
So, tension $T_{2}$ is equal to $10 N$ .

AMU-2006

LAWS OF MOTION (ADDITIONAL)

372242 The apparent weight of a person in a lift moving downwards is half his apparent weight the same lift moving upwards with the same acceleration. Acceleration of the lift is

1

g

2

\frac{g}{4}

3

\frac{g}{2}

4

\frac{g}{3}

Explanation:

D Let's,
Mass of person $= m$
Acceleration $= a$
During downward and upward moving
Then,
Apparent weight during downward moving $(W_{down})$
$W_{down} = m (g - a)$
Apparent weight during upwards moving $(W_{up})$
$W_{up} = m (g + a)$
Given,
$W_{down} = \frac{1}{2} W_{up}$
Then, $m (g - a) = \frac{1}{2} m (g + a)$
$a = \frac{g}{3}$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372243 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only

2 B only

3 Both A and B

4 neither A nor B

Explanation:

A All masses are pushed by force $F$

Then, acceleration-
$a = \frac{F}{total mass} = \frac{F}{m_{1} + m_{2} + m_{3}} = \frac{F}{3 + 2 + 1}$
$a = \frac{F}{6} m / s^{2}$
So, All block move with same acceleration $\frac{F}{6} m / s^{2}$
Net force on mass, $m_{1} = 3 kg$
$F_{1} = m_{1} \cdot a = 3 \cdot \frac{F}{6} = \frac{F}{2}$
$F_{1} = \frac{F}{2} N$
Net force on mass, $m_{2} = 2 kg$
$F_{2} = m_{2} \cdot a = 2 \cdot \frac{F}{6} = \frac{F}{3}$
$F_{2} = \frac{F}{3} N$
Net force on mass, $m_{3} = 1 kg$
$F_{3} = m_{3} \cdot a = 1 \times \frac{F}{6} = \frac{F}{6}$
$F_{3} = \frac{F}{6} N$
So, the net force on each block are different.

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372240 Two blocks of masses $m$ and $2 m$ are connected by a light string passing over a frictionless pulley. As shown in the figure, the mass $m$ is placed on a smooth inclined plane of inclination $30^{\circ}$ and $2 m$ hangs vertically. If the system is released, the blocks move with an acceleration equal to

1

g / 4

2

g / 3

3

g / 2

4

g

Explanation:

C In downward,
$2 mg - T = 2 ma$
And
$mg \sin 30^{\circ} - T = - ma$

Subtracting both equation,
$2 mg - \frac{mg}{2} = 3 ma$
$3 a = \frac{ug - g}{2}$
$3 a = \frac{3 g}{2}$
$a = \frac{g}{2}$

AMU-2009

LAWS OF MOTION (ADDITIONAL)

372241 In the given figure, three blocks are connected by massless cords. Masses of the block are $m$ , $3 m$ ad $5 m$ and they are pulled by a force $F$ on a frictionless horizontal surface. If the tension $T_{1} = 16 N$ , then $T_{2}$ is equal to :

1

6 N

2

10 N

3

(80 / 3) N

4

18 N

Explanation:

B Given, $T_{1} = 16 N$ ,
$m_{1} = m, m_{2} = 3 m, m_{3} = 5 m$
$T_{2} = ?$
Acceleration " $a$ " due to tension $T_{1}$ in mass $m_{2}$ and $m_{3}$
$a = \frac{T_{1}}{m_{2} + m_{3}} = \frac{16}{3 m + 5 m}$
$a = \frac{2}{m}$
Now, tension
$T_{2} = m_{3} \times a$
$T_{2} = 5 m \times \frac{2}{m}$
$T_{2} = 10 N$
So, tension $T_{2}$ is equal to $10 N$ .

AMU-2006

LAWS OF MOTION (ADDITIONAL)

372242 The apparent weight of a person in a lift moving downwards is half his apparent weight the same lift moving upwards with the same acceleration. Acceleration of the lift is

1

g

2

\frac{g}{4}

3

\frac{g}{2}

4

\frac{g}{3}

Explanation:

D Let's,
Mass of person $= m$
Acceleration $= a$
During downward and upward moving
Then,
Apparent weight during downward moving $(W_{down})$
$W_{down} = m (g - a)$
Apparent weight during upwards moving $(W_{up})$
$W_{up} = m (g + a)$
Given,
$W_{down} = \frac{1}{2} W_{up}$
Then, $m (g - a) = \frac{1}{2} m (g + a)$
$a = \frac{g}{3}$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372243 Consider the following statement about the blocks, shown in the diagram, which are being pushed by a constant force on a frictionless table.

1 A only

2 B only

3 Both A and B

4 neither A nor B

Explanation:

A All masses are pushed by force $F$

Then, acceleration-
$a = \frac{F}{total mass} = \frac{F}{m_{1} + m_{2} + m_{3}} = \frac{F}{3 + 2 + 1}$
$a = \frac{F}{6} m / s^{2}$
So, All block move with same acceleration $\frac{F}{6} m / s^{2}$
Net force on mass, $m_{1} = 3 kg$
$F_{1} = m_{1} \cdot a = 3 \cdot \frac{F}{6} = \frac{F}{2}$
$F_{1} = \frac{F}{2} N$
Net force on mass, $m_{2} = 2 kg$
$F_{2} = m_{2} \cdot a = 2 \cdot \frac{F}{6} = \frac{F}{3}$
$F_{2} = \frac{F}{3} N$
Net force on mass, $m_{3} = 1 kg$
$F_{3} = m_{3} \cdot a = 1 \times \frac{F}{6} = \frac{F}{6}$
$F_{3} = \frac{F}{6} N$
So, the net force on each block are different.

AP EMCET(Medical)-2011