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PHXI05:LAWS OF MOTION

363335 A conveyor belt is moving at a constant speed of 2 $m / s$ . A box is gently dropped on it. The coefficient of friction between them is $μ = 0.5$ . The distance that the box will move relative to belt before coming to rest on it, taking $g = 10 m / s^{2}$

1

1.2 m

2

0.4 m

3

Z e r o

4

0.6 m

Explanation:

The frictional force on the box is
$F_{k} = μ_{k} m g$
$μ_{u} m g = m a \Rightarrow a = μ_{u} g$
The velocity of the box w.r.to ground is
$v_{r} & u_{r}$ are relative velocities of the box w.r. to belt
$u_{r} = 2 m / s, v_{r} = 0$
$a = - μ_{u} g$
$\Rightarrow s = \frac{2}{μ_{u} g} = \frac{2}{5} m$
$= 0.4 m$

PHXI05:LAWS OF MOTION

363336 A block of mass 5 $k g$ is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 $N$ and 4 $N$ act on this block . The acceleration of the block is ( $g = 10 m / s^{2}$ )

1

z e r o

2

0.1 m / s^{2}

3

0.2 m / s^{2}

4

0.3 m / s^{2}

Explanation:

Net external force $F = \sqrt{4^{2} + 3^{2}} = 5 N$
Maximum friction
$f_{max} = μ m g = (0.09) (5) (10) = 4.5 N$
Since $F > f_{max}$ , block will move with an acceleration
$a = \frac{F - f_{max}}{m} = \frac{5 - 4.5}{5} = 0.1 m / s^{2}$

PHXI05:LAWS OF MOTION

363337 Three blocks $m_{1}, m_{2}$ , and $m_{3}$ of masses $1.5 k g, 2.0 k g$ , and $10 k g,$ respectively, are placed on a rough surface $(μ = 0.20)$ , as shown in the figure. If a force $F$ is applied to give blocks an acceleration of $3.0 m / s^{2}$ , then the force that the $1.50 k g$ block exerts on the $2.0 k g$ block will be approximately
supporting img

1

10 N

2

15 N

3

17 N

4

12 N

Explanation:

For block of mass the frictional force is
$f_{1} = 0 \cdot 2 \times 1 \cdot 5 \times 10 = 3 N$
Let $N_{1}$ be the force exerted by $m_{2}$ on $m_{1}$ .
$∴ F - 3 - N_{1} = m_{1} a$
$\Rightarrow F - 3 - N_{1} = 1.5 \times 3$
$\Rightarrow F - 3 - N_{1} = 4.5 (1)$
Also, $F - (f_{1} + f_{2} + f_{3}) = (m_{1} + m_{2} + m_{3}) a$
$F - [3 + 4 + 2] = 4.5 \times 3$
$\Rightarrow F = 22.5 N$
$∴$ From Eq. (1)
$22 \cdot 5 - 3 - N_{1} = 4.5, \Rightarrow N_{1} = 15 N$

PHXI05:LAWS OF MOTION

363338 A 10 $k g$ block is resting on the horizontal surface when the force $F$ is applied to it for 7 sec. The variation of $F$ with time is shown in the figure. The coefficients of static and kinetic friction both are equal to 0.50. Find velocity of the block (in $m$ / $s$ ) at $t = 4 \sec$ .
supporting img

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

The force on the block as a function of time is $F = 25 t$
At $t = 2 s$ the block starts moving as $F = f$ .
The net force acting the block is $F_{N e t} = 25 t - 50$
The impulse $m V = \int_{1}^{4} (25 t - 50) d t$
$m = 10 k g \Rightarrow V = 5 m / s$

PHXI05:LAWS OF MOTION

363335 A conveyor belt is moving at a constant speed of 2 $m / s$ . A box is gently dropped on it. The coefficient of friction between them is $μ = 0.5$ . The distance that the box will move relative to belt before coming to rest on it, taking $g = 10 m / s^{2}$

1

1.2 m

2

0.4 m

3

Z e r o

4

0.6 m

Explanation:

The frictional force on the box is
$F_{k} = μ_{k} m g$
$μ_{u} m g = m a \Rightarrow a = μ_{u} g$
The velocity of the box w.r.to ground is
$v_{r} & u_{r}$ are relative velocities of the box w.r. to belt
$u_{r} = 2 m / s, v_{r} = 0$
$a = - μ_{u} g$
$\Rightarrow s = \frac{2}{μ_{u} g} = \frac{2}{5} m$
$= 0.4 m$

PHXI05:LAWS OF MOTION

363336 A block of mass 5 $k g$ is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 $N$ and 4 $N$ act on this block . The acceleration of the block is ( $g = 10 m / s^{2}$ )

1

z e r o

2

0.1 m / s^{2}

3

0.2 m / s^{2}

4

0.3 m / s^{2}

Explanation:

Net external force $F = \sqrt{4^{2} + 3^{2}} = 5 N$
Maximum friction
$f_{max} = μ m g = (0.09) (5) (10) = 4.5 N$
Since $F > f_{max}$ , block will move with an acceleration
$a = \frac{F - f_{max}}{m} = \frac{5 - 4.5}{5} = 0.1 m / s^{2}$

PHXI05:LAWS OF MOTION

363337 Three blocks $m_{1}, m_{2}$ , and $m_{3}$ of masses $1.5 k g, 2.0 k g$ , and $10 k g,$ respectively, are placed on a rough surface $(μ = 0.20)$ , as shown in the figure. If a force $F$ is applied to give blocks an acceleration of $3.0 m / s^{2}$ , then the force that the $1.50 k g$ block exerts on the $2.0 k g$ block will be approximately
supporting img

1

10 N

2

15 N

3

17 N

4

12 N

Explanation:

For block of mass the frictional force is
$f_{1} = 0 \cdot 2 \times 1 \cdot 5 \times 10 = 3 N$
Let $N_{1}$ be the force exerted by $m_{2}$ on $m_{1}$ .
$∴ F - 3 - N_{1} = m_{1} a$
$\Rightarrow F - 3 - N_{1} = 1.5 \times 3$
$\Rightarrow F - 3 - N_{1} = 4.5 (1)$
Also, $F - (f_{1} + f_{2} + f_{3}) = (m_{1} + m_{2} + m_{3}) a$
$F - [3 + 4 + 2] = 4.5 \times 3$
$\Rightarrow F = 22.5 N$
$∴$ From Eq. (1)
$22 \cdot 5 - 3 - N_{1} = 4.5, \Rightarrow N_{1} = 15 N$

PHXI05:LAWS OF MOTION

363338 A 10 $k g$ block is resting on the horizontal surface when the force $F$ is applied to it for 7 sec. The variation of $F$ with time is shown in the figure. The coefficients of static and kinetic friction both are equal to 0.50. Find velocity of the block (in $m$ / $s$ ) at $t = 4 \sec$ .
supporting img

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

The force on the block as a function of time is $F = 25 t$
At $t = 2 s$ the block starts moving as $F = f$ .
The net force acting the block is $F_{N e t} = 25 t - 50$
The impulse $m V = \int_{1}^{4} (25 t - 50) d t$
$m = 10 k g \Rightarrow V = 5 m / s$

PHXI05:LAWS OF MOTION

363335 A conveyor belt is moving at a constant speed of 2 $m / s$ . A box is gently dropped on it. The coefficient of friction between them is $μ = 0.5$ . The distance that the box will move relative to belt before coming to rest on it, taking $g = 10 m / s^{2}$

1

1.2 m

2

0.4 m

3

Z e r o

4

0.6 m

Explanation:

The frictional force on the box is
$F_{k} = μ_{k} m g$
$μ_{u} m g = m a \Rightarrow a = μ_{u} g$
The velocity of the box w.r.to ground is
$v_{r} & u_{r}$ are relative velocities of the box w.r. to belt
$u_{r} = 2 m / s, v_{r} = 0$
$a = - μ_{u} g$
$\Rightarrow s = \frac{2}{μ_{u} g} = \frac{2}{5} m$
$= 0.4 m$

PHXI05:LAWS OF MOTION

363336 A block of mass 5 $k g$ is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 $N$ and 4 $N$ act on this block . The acceleration of the block is ( $g = 10 m / s^{2}$ )

1

z e r o

2

0.1 m / s^{2}

3

0.2 m / s^{2}

4

0.3 m / s^{2}

Explanation:

Net external force $F = \sqrt{4^{2} + 3^{2}} = 5 N$
Maximum friction
$f_{max} = μ m g = (0.09) (5) (10) = 4.5 N$
Since $F > f_{max}$ , block will move with an acceleration
$a = \frac{F - f_{max}}{m} = \frac{5 - 4.5}{5} = 0.1 m / s^{2}$

PHXI05:LAWS OF MOTION

363337 Three blocks $m_{1}, m_{2}$ , and $m_{3}$ of masses $1.5 k g, 2.0 k g$ , and $10 k g,$ respectively, are placed on a rough surface $(μ = 0.20)$ , as shown in the figure. If a force $F$ is applied to give blocks an acceleration of $3.0 m / s^{2}$ , then the force that the $1.50 k g$ block exerts on the $2.0 k g$ block will be approximately
supporting img

1

10 N

2

15 N

3

17 N

4

12 N

Explanation:

For block of mass the frictional force is
$f_{1} = 0 \cdot 2 \times 1 \cdot 5 \times 10 = 3 N$
Let $N_{1}$ be the force exerted by $m_{2}$ on $m_{1}$ .
$∴ F - 3 - N_{1} = m_{1} a$
$\Rightarrow F - 3 - N_{1} = 1.5 \times 3$
$\Rightarrow F - 3 - N_{1} = 4.5 (1)$
Also, $F - (f_{1} + f_{2} + f_{3}) = (m_{1} + m_{2} + m_{3}) a$
$F - [3 + 4 + 2] = 4.5 \times 3$
$\Rightarrow F = 22.5 N$
$∴$ From Eq. (1)
$22 \cdot 5 - 3 - N_{1} = 4.5, \Rightarrow N_{1} = 15 N$

PHXI05:LAWS OF MOTION

363338 A 10 $k g$ block is resting on the horizontal surface when the force $F$ is applied to it for 7 sec. The variation of $F$ with time is shown in the figure. The coefficients of static and kinetic friction both are equal to 0.50. Find velocity of the block (in $m$ / $s$ ) at $t = 4 \sec$ .
supporting img

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

The force on the block as a function of time is $F = 25 t$
At $t = 2 s$ the block starts moving as $F = f$ .
The net force acting the block is $F_{N e t} = 25 t - 50$
The impulse $m V = \int_{1}^{4} (25 t - 50) d t$
$m = 10 k g \Rightarrow V = 5 m / s$

PHXI05:LAWS OF MOTION

363335 A conveyor belt is moving at a constant speed of 2 $m / s$ . A box is gently dropped on it. The coefficient of friction between them is $μ = 0.5$ . The distance that the box will move relative to belt before coming to rest on it, taking $g = 10 m / s^{2}$

1

1.2 m

2

0.4 m

3

Z e r o

4

0.6 m

Explanation:

The frictional force on the box is
$F_{k} = μ_{k} m g$
$μ_{u} m g = m a \Rightarrow a = μ_{u} g$
The velocity of the box w.r.to ground is
$v_{r} & u_{r}$ are relative velocities of the box w.r. to belt
$u_{r} = 2 m / s, v_{r} = 0$
$a = - μ_{u} g$
$\Rightarrow s = \frac{2}{μ_{u} g} = \frac{2}{5} m$
$= 0.4 m$

PHXI05:LAWS OF MOTION

363336 A block of mass 5 $k g$ is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 $N$ and 4 $N$ act on this block . The acceleration of the block is ( $g = 10 m / s^{2}$ )

1

z e r o

2

0.1 m / s^{2}

3

0.2 m / s^{2}

4

0.3 m / s^{2}

Explanation:

Net external force $F = \sqrt{4^{2} + 3^{2}} = 5 N$
Maximum friction
$f_{max} = μ m g = (0.09) (5) (10) = 4.5 N$
Since $F > f_{max}$ , block will move with an acceleration
$a = \frac{F - f_{max}}{m} = \frac{5 - 4.5}{5} = 0.1 m / s^{2}$

PHXI05:LAWS OF MOTION

363337 Three blocks $m_{1}, m_{2}$ , and $m_{3}$ of masses $1.5 k g, 2.0 k g$ , and $10 k g,$ respectively, are placed on a rough surface $(μ = 0.20)$ , as shown in the figure. If a force $F$ is applied to give blocks an acceleration of $3.0 m / s^{2}$ , then the force that the $1.50 k g$ block exerts on the $2.0 k g$ block will be approximately
supporting img

1

10 N

2

15 N

3

17 N

4

12 N

Explanation:

For block of mass the frictional force is
$f_{1} = 0 \cdot 2 \times 1 \cdot 5 \times 10 = 3 N$
Let $N_{1}$ be the force exerted by $m_{2}$ on $m_{1}$ .
$∴ F - 3 - N_{1} = m_{1} a$
$\Rightarrow F - 3 - N_{1} = 1.5 \times 3$
$\Rightarrow F - 3 - N_{1} = 4.5 (1)$
Also, $F - (f_{1} + f_{2} + f_{3}) = (m_{1} + m_{2} + m_{3}) a$
$F - [3 + 4 + 2] = 4.5 \times 3$
$\Rightarrow F = 22.5 N$
$∴$ From Eq. (1)
$22 \cdot 5 - 3 - N_{1} = 4.5, \Rightarrow N_{1} = 15 N$

PHXI05:LAWS OF MOTION

363338 A 10 $k g$ block is resting on the horizontal surface when the force $F$ is applied to it for 7 sec. The variation of $F$ with time is shown in the figure. The coefficients of static and kinetic friction both are equal to 0.50. Find velocity of the block (in $m$ / $s$ ) at $t = 4 \sec$ .
supporting img

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

The force on the block as a function of time is $F = 25 t$
At $t = 2 s$ the block starts moving as $F = f$ .
The net force acting the block is $F_{N e t} = 25 t - 50$
The impulse $m V = \int_{1}^{4} (25 t - 50) d t$
$m = 10 k g \Rightarrow V = 5 m / s$

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