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Laws of Motion

270356 A man of mass $m$ stands on a platform of equal mass $m$ and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be
![original image](https://cdn.mathpix.com/snip/images/cFlqGO3lAJncD5Y-VEHeP6ooxNU6sFZ6ZZTeuDXo6EU.original.fullsize.png)

1

\frac{g}{2}

2

\frac{g}{4}

3

g

4 zero

Explanation:

$F = 4 ◻ \frac{m g}{2} \to\to$ upwards
$W = 2 m g \to$ downwards
$F = W, ∴ a = 0$

Laws of Motion

270357 A block is sliding along inclined plane as shown in figure. If the accelerationofcham ber is ' $a$ ' as shown in the figure. The time required to cover a distance $L$ along inclined plane is

1

\sqrt{\frac{2 L}{g \sin θ - a \cos θ}}

2

\sqrt{\frac{2 L}{g \sin θ + a \sin θ}}

3

\sqrt{\frac{2 L}{g \sin θ + a \cos θ}}

4

\sqrt{\frac{2 L}{g \sin θ}}

Explanation:

$f_{net} ($ downward $) = m g \sin θ + m a \cos θ$
$= m (g \sin θ + a \cos θ)$
$∴ g_{eff} = g \sin θ + a \cos θ; T = \sqrt{\frac{2 L}{g_{eff}}}$

Laws of Motion

270358 An inclined plane makes an angle $30^{\circ}$ with the horizontal. A groove (OA) of length $5 m$ cut, in the plane makes an angle $30^{\circ}$ with $O X$ . A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to $O$ is $(g = 10 m / s^{2})$

1

4 s

2

2 s

3

3 s

4

1 s

Explanation:

Acceleration of the cylinder down the plane is
$a = (g \sin θ_{1}) (\sin θ_{2}); t = \sqrt{\frac{2 s}{a}}$

Laws of Motion

270359 Two masses each equal to $m$ are lying on $X$ axis at $(- a, 0)$ and $(+ a, 0)$ , respectively, as shown in fig. They are connected by a light string. A force $F$ is applied at the origin along vertical direction. As a result, the masses move towards each other without loosing contact with ground. What is the acceleration of each mass? Assume the instantaneous position of the masses as $(- x, 0)$ and $(x, 0)$ , respectively

1

\frac{2 F}{m} \frac{\sqrt{(a^{2} - x^{2})}}{x}

2

\frac{2 F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

3

\frac{F}{2 m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

4

\frac{F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

Explanation:

$F = 2 T \sin θ; m a^{1} = T \cos θ$

Laws of Motion

270356 A man of mass $m$ stands on a platform of equal mass $m$ and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be
![original image](https://cdn.mathpix.com/snip/images/cFlqGO3lAJncD5Y-VEHeP6ooxNU6sFZ6ZZTeuDXo6EU.original.fullsize.png)

1

\frac{g}{2}

2

\frac{g}{4}

3

g

4 zero

Explanation:

$F = 4 ◻ \frac{m g}{2} \to\to$ upwards
$W = 2 m g \to$ downwards
$F = W, ∴ a = 0$

Laws of Motion

270357 A block is sliding along inclined plane as shown in figure. If the accelerationofcham ber is ' $a$ ' as shown in the figure. The time required to cover a distance $L$ along inclined plane is

1

\sqrt{\frac{2 L}{g \sin θ - a \cos θ}}

2

\sqrt{\frac{2 L}{g \sin θ + a \sin θ}}

3

\sqrt{\frac{2 L}{g \sin θ + a \cos θ}}

4

\sqrt{\frac{2 L}{g \sin θ}}

Explanation:

$f_{net} ($ downward $) = m g \sin θ + m a \cos θ$
$= m (g \sin θ + a \cos θ)$
$∴ g_{eff} = g \sin θ + a \cos θ; T = \sqrt{\frac{2 L}{g_{eff}}}$

Laws of Motion

270358 An inclined plane makes an angle $30^{\circ}$ with the horizontal. A groove (OA) of length $5 m$ cut, in the plane makes an angle $30^{\circ}$ with $O X$ . A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to $O$ is $(g = 10 m / s^{2})$

1

4 s

2

2 s

3

3 s

4

1 s

Explanation:

Acceleration of the cylinder down the plane is
$a = (g \sin θ_{1}) (\sin θ_{2}); t = \sqrt{\frac{2 s}{a}}$

Laws of Motion

270359 Two masses each equal to $m$ are lying on $X$ axis at $(- a, 0)$ and $(+ a, 0)$ , respectively, as shown in fig. They are connected by a light string. A force $F$ is applied at the origin along vertical direction. As a result, the masses move towards each other without loosing contact with ground. What is the acceleration of each mass? Assume the instantaneous position of the masses as $(- x, 0)$ and $(x, 0)$ , respectively

1

\frac{2 F}{m} \frac{\sqrt{(a^{2} - x^{2})}}{x}

2

\frac{2 F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

3

\frac{F}{2 m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

4

\frac{F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

Explanation:

$F = 2 T \sin θ; m a^{1} = T \cos θ$

Laws of Motion

270356 A man of mass $m$ stands on a platform of equal mass $m$ and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be
![original image](https://cdn.mathpix.com/snip/images/cFlqGO3lAJncD5Y-VEHeP6ooxNU6sFZ6ZZTeuDXo6EU.original.fullsize.png)

1

\frac{g}{2}

2

\frac{g}{4}

3

g

4 zero

Explanation:

$F = 4 ◻ \frac{m g}{2} \to\to$ upwards
$W = 2 m g \to$ downwards
$F = W, ∴ a = 0$

Laws of Motion

270357 A block is sliding along inclined plane as shown in figure. If the accelerationofcham ber is ' $a$ ' as shown in the figure. The time required to cover a distance $L$ along inclined plane is

1

\sqrt{\frac{2 L}{g \sin θ - a \cos θ}}

2

\sqrt{\frac{2 L}{g \sin θ + a \sin θ}}

3

\sqrt{\frac{2 L}{g \sin θ + a \cos θ}}

4

\sqrt{\frac{2 L}{g \sin θ}}

Explanation:

$f_{net} ($ downward $) = m g \sin θ + m a \cos θ$
$= m (g \sin θ + a \cos θ)$
$∴ g_{eff} = g \sin θ + a \cos θ; T = \sqrt{\frac{2 L}{g_{eff}}}$

Laws of Motion

270358 An inclined plane makes an angle $30^{\circ}$ with the horizontal. A groove (OA) of length $5 m$ cut, in the plane makes an angle $30^{\circ}$ with $O X$ . A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to $O$ is $(g = 10 m / s^{2})$

1

4 s

2

2 s

3

3 s

4

1 s

Explanation:

Acceleration of the cylinder down the plane is
$a = (g \sin θ_{1}) (\sin θ_{2}); t = \sqrt{\frac{2 s}{a}}$

Laws of Motion

270359 Two masses each equal to $m$ are lying on $X$ axis at $(- a, 0)$ and $(+ a, 0)$ , respectively, as shown in fig. They are connected by a light string. A force $F$ is applied at the origin along vertical direction. As a result, the masses move towards each other without loosing contact with ground. What is the acceleration of each mass? Assume the instantaneous position of the masses as $(- x, 0)$ and $(x, 0)$ , respectively

1

\frac{2 F}{m} \frac{\sqrt{(a^{2} - x^{2})}}{x}

2

\frac{2 F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

3

\frac{F}{2 m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

4

\frac{F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

Explanation:

$F = 2 T \sin θ; m a^{1} = T \cos θ$

Laws of Motion

270356 A man of mass $m$ stands on a platform of equal mass $m$ and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be
![original image](https://cdn.mathpix.com/snip/images/cFlqGO3lAJncD5Y-VEHeP6ooxNU6sFZ6ZZTeuDXo6EU.original.fullsize.png)

1

\frac{g}{2}

2

\frac{g}{4}

3

g

4 zero

Explanation:

$F = 4 ◻ \frac{m g}{2} \to\to$ upwards
$W = 2 m g \to$ downwards
$F = W, ∴ a = 0$

Laws of Motion

270357 A block is sliding along inclined plane as shown in figure. If the accelerationofcham ber is ' $a$ ' as shown in the figure. The time required to cover a distance $L$ along inclined plane is

1

\sqrt{\frac{2 L}{g \sin θ - a \cos θ}}

2

\sqrt{\frac{2 L}{g \sin θ + a \sin θ}}

3

\sqrt{\frac{2 L}{g \sin θ + a \cos θ}}

4

\sqrt{\frac{2 L}{g \sin θ}}

Explanation:

$f_{net} ($ downward $) = m g \sin θ + m a \cos θ$
$= m (g \sin θ + a \cos θ)$
$∴ g_{eff} = g \sin θ + a \cos θ; T = \sqrt{\frac{2 L}{g_{eff}}}$

Laws of Motion

270358 An inclined plane makes an angle $30^{\circ}$ with the horizontal. A groove (OA) of length $5 m$ cut, in the plane makes an angle $30^{\circ}$ with $O X$ . A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to $O$ is $(g = 10 m / s^{2})$

1

4 s

2

2 s

3

3 s

4

1 s

Explanation:

Acceleration of the cylinder down the plane is
$a = (g \sin θ_{1}) (\sin θ_{2}); t = \sqrt{\frac{2 s}{a}}$

Laws of Motion

270359 Two masses each equal to $m$ are lying on $X$ axis at $(- a, 0)$ and $(+ a, 0)$ , respectively, as shown in fig. They are connected by a light string. A force $F$ is applied at the origin along vertical direction. As a result, the masses move towards each other without loosing contact with ground. What is the acceleration of each mass? Assume the instantaneous position of the masses as $(- x, 0)$ and $(x, 0)$ , respectively

1

\frac{2 F}{m} \frac{\sqrt{(a^{2} - x^{2})}}{x}

2

\frac{2 F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

3

\frac{F}{2 m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

4

\frac{F}{m} \frac{x}{\sqrt{(a^{2} - x^{2})}}

Explanation:

$F = 2 T \sin θ; m a^{1} = T \cos θ$

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