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LAWS OF MOTION (ADDITIONAL)

372133 Block $B$ lying on a table weighs $W$ . The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B Given,
weight of block $B = W$

Assume, weight of A is $W^{'}$ .
F.B.D for equilibrium of the system

$T \cos θ = μ N = μ W$
$T \sin θ = W^{'}$
[Where $T =$ tension between knot and support] Now, eq.(ii) $\div$ eq.(i), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{W^{'}}{μ W}$
$\Rightarrow \tan θ = \frac{W^{'}}{μ W}$
$\Rightarrow W^{'} = μ W \tan θ$
Hence, the maximum weight of block $A (W^{'}) = μ W \tan θ$

WB JEE 2015

LAWS OF MOTION (ADDITIONAL)

372134 A box of mass $2 kg$ is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and $g = 10 {ms}^{- 2}$ .
The maximum acceleration of the car, for the box to remain stationary, is

1

8 {ms}^{- 2}

2

6 {ms}^{- 2}

3

4 {ms}^{- 2}

4

2 {ms}^{- 2}

Explanation:

D Given,
Mass of box $(m) = 2 kg$

Coefficient of static friction $(μ) = 0.2$
Here, $\sum F_{x} = 0$
For box remain in stationary
Applied force $=$ friction force
$F = f_{r}$
$m a = μ . N$
$m a = μ . mg$
$\Rightarrow a = μ g$
$\Rightarrow a = 0.2 \times 10$
$\Rightarrow a = 2 m / \sec^{2}$
Hence, maximum acceleration (a) $= 2 m / s^{2}$ for the box remain in stationary

WB JEE 2012

LAWS OF MOTION (ADDITIONAL)

372135 Force required to move a mass of $1 kg$ at rest on a horizontal rough plane $(μ = 0.1$ and $g = 9.8$ $m / s^{2}$ ) is

1

0.98 N

2

0.49 N

3

9.8 N

4

4.9 N

Explanation:

A
Given, $m = 1 kg, μ = 0.1$
According to question -
Applied force $=$ friction force,
$F = f_{r}$
$F = μ . N$
$F = μ . mg$
$F = 0.1 \times 1 \times 9.8$
$F = 0.98 N$

WB JEE 2008

LAWS OF MOTION (ADDITIONAL)

372136 A block kept on a rough surface starts sliding when the inclination of the surface is $θ$ with respect to the horizontal. The coefficient of static friction between the block and the surface is :

1

\sec θ

2

\sin θ

3

\tan θ

4

\cos θ

Explanation:

C As the block starts slipping for angle of inclination $θ$ , it must be equal to angle of repose given by

According to figure -
$N = mg \cos θ$
and, $f = μ N = mg \sin θ$
From equation (i) and (ii)
$\frac{μ N}{N} = \frac{mg \sin θ}{mg \cos θ}$
$μ_{s} = \tan θ$

Karnataka CET-2012

LAWS OF MOTION (ADDITIONAL)

372133 Block $B$ lying on a table weighs $W$ . The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B Given,
weight of block $B = W$

Assume, weight of A is $W^{'}$ .
F.B.D for equilibrium of the system

$T \cos θ = μ N = μ W$
$T \sin θ = W^{'}$
[Where $T =$ tension between knot and support] Now, eq.(ii) $\div$ eq.(i), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{W^{'}}{μ W}$
$\Rightarrow \tan θ = \frac{W^{'}}{μ W}$
$\Rightarrow W^{'} = μ W \tan θ$
Hence, the maximum weight of block $A (W^{'}) = μ W \tan θ$

WB JEE 2015

LAWS OF MOTION (ADDITIONAL)

372134 A box of mass $2 kg$ is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and $g = 10 {ms}^{- 2}$ .
The maximum acceleration of the car, for the box to remain stationary, is

1

8 {ms}^{- 2}

2

6 {ms}^{- 2}

3

4 {ms}^{- 2}

4

2 {ms}^{- 2}

Explanation:

D Given,
Mass of box $(m) = 2 kg$

Coefficient of static friction $(μ) = 0.2$
Here, $\sum F_{x} = 0$
For box remain in stationary
Applied force $=$ friction force
$F = f_{r}$
$m a = μ . N$
$m a = μ . mg$
$\Rightarrow a = μ g$
$\Rightarrow a = 0.2 \times 10$
$\Rightarrow a = 2 m / \sec^{2}$
Hence, maximum acceleration (a) $= 2 m / s^{2}$ for the box remain in stationary

WB JEE 2012

LAWS OF MOTION (ADDITIONAL)

372135 Force required to move a mass of $1 kg$ at rest on a horizontal rough plane $(μ = 0.1$ and $g = 9.8$ $m / s^{2}$ ) is

1

0.98 N

2

0.49 N

3

9.8 N

4

4.9 N

Explanation:

A
Given, $m = 1 kg, μ = 0.1$
According to question -
Applied force $=$ friction force,
$F = f_{r}$
$F = μ . N$
$F = μ . mg$
$F = 0.1 \times 1 \times 9.8$
$F = 0.98 N$

WB JEE 2008

LAWS OF MOTION (ADDITIONAL)

372136 A block kept on a rough surface starts sliding when the inclination of the surface is $θ$ with respect to the horizontal. The coefficient of static friction between the block and the surface is :

1

\sec θ

2

\sin θ

3

\tan θ

4

\cos θ

Explanation:

C As the block starts slipping for angle of inclination $θ$ , it must be equal to angle of repose given by

According to figure -
$N = mg \cos θ$
and, $f = μ N = mg \sin θ$
From equation (i) and (ii)
$\frac{μ N}{N} = \frac{mg \sin θ}{mg \cos θ}$
$μ_{s} = \tan θ$

Karnataka CET-2012

LAWS OF MOTION (ADDITIONAL)

372133 Block $B$ lying on a table weighs $W$ . The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B Given,
weight of block $B = W$

Assume, weight of A is $W^{'}$ .
F.B.D for equilibrium of the system

$T \cos θ = μ N = μ W$
$T \sin θ = W^{'}$
[Where $T =$ tension between knot and support] Now, eq.(ii) $\div$ eq.(i), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{W^{'}}{μ W}$
$\Rightarrow \tan θ = \frac{W^{'}}{μ W}$
$\Rightarrow W^{'} = μ W \tan θ$
Hence, the maximum weight of block $A (W^{'}) = μ W \tan θ$

WB JEE 2015

LAWS OF MOTION (ADDITIONAL)

372134 A box of mass $2 kg$ is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and $g = 10 {ms}^{- 2}$ .
The maximum acceleration of the car, for the box to remain stationary, is

1

8 {ms}^{- 2}

2

6 {ms}^{- 2}

3

4 {ms}^{- 2}

4

2 {ms}^{- 2}

Explanation:

D Given,
Mass of box $(m) = 2 kg$

Coefficient of static friction $(μ) = 0.2$
Here, $\sum F_{x} = 0$
For box remain in stationary
Applied force $=$ friction force
$F = f_{r}$
$m a = μ . N$
$m a = μ . mg$
$\Rightarrow a = μ g$
$\Rightarrow a = 0.2 \times 10$
$\Rightarrow a = 2 m / \sec^{2}$
Hence, maximum acceleration (a) $= 2 m / s^{2}$ for the box remain in stationary

WB JEE 2012

LAWS OF MOTION (ADDITIONAL)

372135 Force required to move a mass of $1 kg$ at rest on a horizontal rough plane $(μ = 0.1$ and $g = 9.8$ $m / s^{2}$ ) is

1

0.98 N

2

0.49 N

3

9.8 N

4

4.9 N

Explanation:

A
Given, $m = 1 kg, μ = 0.1$
According to question -
Applied force $=$ friction force,
$F = f_{r}$
$F = μ . N$
$F = μ . mg$
$F = 0.1 \times 1 \times 9.8$
$F = 0.98 N$

WB JEE 2008

LAWS OF MOTION (ADDITIONAL)

372136 A block kept on a rough surface starts sliding when the inclination of the surface is $θ$ with respect to the horizontal. The coefficient of static friction between the block and the surface is :

1

\sec θ

2

\sin θ

3

\tan θ

4

\cos θ

Explanation:

C As the block starts slipping for angle of inclination $θ$ , it must be equal to angle of repose given by

According to figure -
$N = mg \cos θ$
and, $f = μ N = mg \sin θ$
From equation (i) and (ii)
$\frac{μ N}{N} = \frac{mg \sin θ}{mg \cos θ}$
$μ_{s} = \tan θ$

Karnataka CET-2012

LAWS OF MOTION (ADDITIONAL)

372133 Block $B$ lying on a table weighs $W$ . The coefficient of static friction between the block and the table is $μ$ . Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

1

\frac{W \tan θ}{μ}

2

μ W \tan θ

3

μ W \sqrt{1 + \tan^{2} θ}

4

μ W \sin θ

Explanation:

B Given,
weight of block $B = W$

Assume, weight of A is $W^{'}$ .
F.B.D for equilibrium of the system

$T \cos θ = μ N = μ W$
$T \sin θ = W^{'}$
[Where $T =$ tension between knot and support] Now, eq.(ii) $\div$ eq.(i), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{W^{'}}{μ W}$
$\Rightarrow \tan θ = \frac{W^{'}}{μ W}$
$\Rightarrow W^{'} = μ W \tan θ$
Hence, the maximum weight of block $A (W^{'}) = μ W \tan θ$

WB JEE 2015

LAWS OF MOTION (ADDITIONAL)

372134 A box of mass $2 kg$ is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and $g = 10 {ms}^{- 2}$ .
The maximum acceleration of the car, for the box to remain stationary, is

1

8 {ms}^{- 2}

2

6 {ms}^{- 2}

3

4 {ms}^{- 2}

4

2 {ms}^{- 2}

Explanation:

D Given,
Mass of box $(m) = 2 kg$

Coefficient of static friction $(μ) = 0.2$
Here, $\sum F_{x} = 0$
For box remain in stationary
Applied force $=$ friction force
$F = f_{r}$
$m a = μ . N$
$m a = μ . mg$
$\Rightarrow a = μ g$
$\Rightarrow a = 0.2 \times 10$
$\Rightarrow a = 2 m / \sec^{2}$
Hence, maximum acceleration (a) $= 2 m / s^{2}$ for the box remain in stationary

WB JEE 2012

LAWS OF MOTION (ADDITIONAL)

372135 Force required to move a mass of $1 kg$ at rest on a horizontal rough plane $(μ = 0.1$ and $g = 9.8$ $m / s^{2}$ ) is

1

0.98 N

2

0.49 N

3

9.8 N

4

4.9 N

Explanation:

A
Given, $m = 1 kg, μ = 0.1$
According to question -
Applied force $=$ friction force,
$F = f_{r}$
$F = μ . N$
$F = μ . mg$
$F = 0.1 \times 1 \times 9.8$
$F = 0.98 N$

WB JEE 2008

LAWS OF MOTION (ADDITIONAL)

372136 A block kept on a rough surface starts sliding when the inclination of the surface is $θ$ with respect to the horizontal. The coefficient of static friction between the block and the surface is :

1

\sec θ

2

\sin θ

3

\tan θ

4

\cos θ

Explanation:

C As the block starts slipping for angle of inclination $θ$ , it must be equal to angle of repose given by

According to figure -
$N = mg \cos θ$
and, $f = μ N = mg \sin θ$
From equation (i) and (ii)
$\frac{μ N}{N} = \frac{mg \sin θ}{mg \cos θ}$
$μ_{s} = \tan θ$

Karnataka CET-2012