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Laws of Motion

146389 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

146391 The weight of a man in a lift, moving upwards is $608 N$ while the weight of the same man in the lift moving downwards, with the same acceleration, is $368 N$ . His normal weight in newton is:

1 488

2 588

3 480

4 240

Explanation:

A Given,
$F_{up} = 608 N, F_{down} = 368 N$
In upward direction
$F_{up} = m (g + a)$
$608 = m (g + a)$
In downward direction
$F_{down} = m (g - a)$
$368 = m (g - a)$
Adding both equation (i) and (ii),
$976 = 2 mg$
$mg = \frac{976}{2} = 488 N$
Normal weight $(W) = mg = 488 N$

AP EAMCET(Medical)-1997

Laws of Motion

146392 In the system shown in the adjoining figure, the tension $T_{2}$ is

1

g

2

2 g

3

5 g

4

6 g

Explanation:

C Given,

\therefore \quad \mathrm{T}_
$\begin{array}{r} T_{2} = (m_{2} + m_{3}) \times g \\ T_{2} = (3 + 2) \times g \\ T_{2} = 5 g \end{array}$

UPSEE - 2012

Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F ⟵ M \overset{T_{1}}{⟵} M \overset{T_{2}}{⟵} M \overset{T_{3}}{⟵} M$

1

T_{1} = \frac{1}{4} F, T_{2} = \frac{3}{2} F, T_{3} = \frac{1}{4} F

2

T_{1} = \frac{1}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

3

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{4} F

4

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

Explanation:

C Given,
$m_{1} = m_{2} = m_{3} = m_{4} = M$
$F ⟵ \underset{m_{1}}{⟵} \underset{m_{2}}{\overset{T_{1}}{⟵}} M \underset{m_{3}}{\overset{T_{2}}{⟵}} M \overset{T_{3}}{\overset{T_{4}}{⟵}} M$
$F = total mass \times acceleration$
$F = (m_{1} + m_{2} + m_{3} + m_{4}) \times a$
$F = (M + M + M + M) \times a$
$a = \frac{F}{4 M}$
$T_{1} = (m_{2} + m_{3} + m_{4}) \times a$
$T_{1} = 3 M \times a = 3 M \times \frac{F}{4 M} = \frac{3}{4} F$
$T_{2} = (m_{3} + m_{4}) \times a$
$T_{2} = 2 M \times a = 2 M \times \frac{F}{4 M} = \frac{1}{2} F$
$T_{3} = m_{4} \times a$
$T_{3} = M \times a = \frac{1}{4} F$

UPSEE - 2009

Laws of Motion

146389 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

146391 The weight of a man in a lift, moving upwards is $608 N$ while the weight of the same man in the lift moving downwards, with the same acceleration, is $368 N$ . His normal weight in newton is:

1 488

2 588

3 480

4 240

Explanation:

A Given,
$F_{up} = 608 N, F_{down} = 368 N$
In upward direction
$F_{up} = m (g + a)$
$608 = m (g + a)$
In downward direction
$F_{down} = m (g - a)$
$368 = m (g - a)$
Adding both equation (i) and (ii),
$976 = 2 mg$
$mg = \frac{976}{2} = 488 N$
Normal weight $(W) = mg = 488 N$

AP EAMCET(Medical)-1997

Laws of Motion

146392 In the system shown in the adjoining figure, the tension $T_{2}$ is

1

g

2

2 g

3

5 g

4

6 g

Explanation:

C Given,

\therefore \quad \mathrm{T}_
$\begin{array}{r} T_{2} = (m_{2} + m_{3}) \times g \\ T_{2} = (3 + 2) \times g \\ T_{2} = 5 g \end{array}$

UPSEE - 2012

Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F ⟵ M \overset{T_{1}}{⟵} M \overset{T_{2}}{⟵} M \overset{T_{3}}{⟵} M$

1

T_{1} = \frac{1}{4} F, T_{2} = \frac{3}{2} F, T_{3} = \frac{1}{4} F

2

T_{1} = \frac{1}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

3

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{4} F

4

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

Explanation:

C Given,
$m_{1} = m_{2} = m_{3} = m_{4} = M$
$F ⟵ \underset{m_{1}}{⟵} \underset{m_{2}}{\overset{T_{1}}{⟵}} M \underset{m_{3}}{\overset{T_{2}}{⟵}} M \overset{T_{3}}{\overset{T_{4}}{⟵}} M$
$F = total mass \times acceleration$
$F = (m_{1} + m_{2} + m_{3} + m_{4}) \times a$
$F = (M + M + M + M) \times a$
$a = \frac{F}{4 M}$
$T_{1} = (m_{2} + m_{3} + m_{4}) \times a$
$T_{1} = 3 M \times a = 3 M \times \frac{F}{4 M} = \frac{3}{4} F$
$T_{2} = (m_{3} + m_{4}) \times a$
$T_{2} = 2 M \times a = 2 M \times \frac{F}{4 M} = \frac{1}{2} F$
$T_{3} = m_{4} \times a$
$T_{3} = M \times a = \frac{1}{4} F$

UPSEE - 2009

Laws of Motion

146389 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

146391 The weight of a man in a lift, moving upwards is $608 N$ while the weight of the same man in the lift moving downwards, with the same acceleration, is $368 N$ . His normal weight in newton is:

1 488

2 588

3 480

4 240

Explanation:

A Given,
$F_{up} = 608 N, F_{down} = 368 N$
In upward direction
$F_{up} = m (g + a)$
$608 = m (g + a)$
In downward direction
$F_{down} = m (g - a)$
$368 = m (g - a)$
Adding both equation (i) and (ii),
$976 = 2 mg$
$mg = \frac{976}{2} = 488 N$
Normal weight $(W) = mg = 488 N$

AP EAMCET(Medical)-1997

Laws of Motion

146392 In the system shown in the adjoining figure, the tension $T_{2}$ is

1

g

2

2 g

3

5 g

4

6 g

Explanation:

C Given,

\therefore \quad \mathrm{T}_
$\begin{array}{r} T_{2} = (m_{2} + m_{3}) \times g \\ T_{2} = (3 + 2) \times g \\ T_{2} = 5 g \end{array}$

UPSEE - 2012

Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F ⟵ M \overset{T_{1}}{⟵} M \overset{T_{2}}{⟵} M \overset{T_{3}}{⟵} M$

1

T_{1} = \frac{1}{4} F, T_{2} = \frac{3}{2} F, T_{3} = \frac{1}{4} F

2

T_{1} = \frac{1}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

3

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{4} F

4

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

Explanation:

C Given,
$m_{1} = m_{2} = m_{3} = m_{4} = M$
$F ⟵ \underset{m_{1}}{⟵} \underset{m_{2}}{\overset{T_{1}}{⟵}} M \underset{m_{3}}{\overset{T_{2}}{⟵}} M \overset{T_{3}}{\overset{T_{4}}{⟵}} M$
$F = total mass \times acceleration$
$F = (m_{1} + m_{2} + m_{3} + m_{4}) \times a$
$F = (M + M + M + M) \times a$
$a = \frac{F}{4 M}$
$T_{1} = (m_{2} + m_{3} + m_{4}) \times a$
$T_{1} = 3 M \times a = 3 M \times \frac{F}{4 M} = \frac{3}{4} F$
$T_{2} = (m_{3} + m_{4}) \times a$
$T_{2} = 2 M \times a = 2 M \times \frac{F}{4 M} = \frac{1}{2} F$
$T_{3} = m_{4} \times a$
$T_{3} = M \times a = \frac{1}{4} F$

UPSEE - 2009

Laws of Motion

146389 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

146391 The weight of a man in a lift, moving upwards is $608 N$ while the weight of the same man in the lift moving downwards, with the same acceleration, is $368 N$ . His normal weight in newton is:

1 488

2 588

3 480

4 240

Explanation:

A Given,
$F_{up} = 608 N, F_{down} = 368 N$
In upward direction
$F_{up} = m (g + a)$
$608 = m (g + a)$
In downward direction
$F_{down} = m (g - a)$
$368 = m (g - a)$
Adding both equation (i) and (ii),
$976 = 2 mg$
$mg = \frac{976}{2} = 488 N$
Normal weight $(W) = mg = 488 N$

AP EAMCET(Medical)-1997

Laws of Motion

146392 In the system shown in the adjoining figure, the tension $T_{2}$ is

1

g

2

2 g

3

5 g

4

6 g

Explanation:

C Given,

\therefore \quad \mathrm{T}_
$\begin{array}{r} T_{2} = (m_{2} + m_{3}) \times g \\ T_{2} = (3 + 2) \times g \\ T_{2} = 5 g \end{array}$

UPSEE - 2012

Laws of Motion

146393 Four blocks of same mass connected by cords are pulled by a force $F$ on a smooth horizontal surface, as shown in figure. The tensions $T_{1}, T_{2}$ and $T_{3}$ will be
$F ⟵ M \overset{T_{1}}{⟵} M \overset{T_{2}}{⟵} M \overset{T_{3}}{⟵} M$

1

T_{1} = \frac{1}{4} F, T_{2} = \frac{3}{2} F, T_{3} = \frac{1}{4} F

2

T_{1} = \frac{1}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

3

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{4} F

4

T_{1} = \frac{3}{4} F, T_{2} = \frac{1}{2} F, T_{3} = \frac{1}{2} F

Explanation:

C Given,
$m_{1} = m_{2} = m_{3} = m_{4} = M$
$F ⟵ \underset{m_{1}}{⟵} \underset{m_{2}}{\overset{T_{1}}{⟵}} M \underset{m_{3}}{\overset{T_{2}}{⟵}} M \overset{T_{3}}{\overset{T_{4}}{⟵}} M$
$F = total mass \times acceleration$
$F = (m_{1} + m_{2} + m_{3} + m_{4}) \times a$
$F = (M + M + M + M) \times a$
$a = \frac{F}{4 M}$
$T_{1} = (m_{2} + m_{3} + m_{4}) \times a$
$T_{1} = 3 M \times a = 3 M \times \frac{F}{4 M} = \frac{3}{4} F$
$T_{2} = (m_{3} + m_{4}) \times a$
$T_{2} = 2 M \times a = 2 M \times \frac{F}{4 M} = \frac{1}{2} F$
$T_{3} = m_{4} \times a$
$T_{3} = M \times a = \frac{1}{4} F$

UPSEE - 2009