QBManager

Laws of Motion

146014 In the arrangement shown in figure $a_{1}, a_{2}, a_{3}$ , and $a_{4}$ are the accelerations of masses $m_{1}, m_{2}$ , $m_{3}$ , and $m_{4}$ respectively. Which of the following relation is true for this arrangement?

1

4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0

2

a_{1} + 4 a_{2} + 3 a_{3} + a_{4} = 0

3

a_{1} + 4 a_{2} + 3 a_{3} + 2 a_{4} = 0

4

2 a_{1} + 2 a_{2} + 3 a_{3} + a_{4} = 0

Explanation:

A
Using constraints,
$Σ \vec{T} \cdot \vec{a} = 0$
$- 4 {Ta}_{1} - 2 {Ta}_{2} - {Ta}_{3} - {Ta}_{4} = 0$
$4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

JEE Main-26.06.2022

Laws of Motion

146015 A cylinder of mass $12 kg$ is sliding on plane with an initial velocity $20 {ms}^{- 1}$ . If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of

1

40 m

2

5 m

3

20 m

4

10 m

Explanation:

A Given,
$m = 12 kg$
$u = 20 {ms}^{- 1}$
$μ = 0.5$
We know that
Friction force $(f) = μ N = μ mg [∴ N = mg]$
$= 0.5 \times 12 \times 10$
$f = 60$
$∵ f = - ma$
Acceleration $(a) = - \frac{f}{m} = - \frac{60}{12} = - 5 {ms}^{- 2}$
From Newton's third law of motion,
$v^{2} = u^{2} + 2 a s$
$0 = (20)^{2} + 2 \times (- 5) \times s$
$0 = 400 - 10 s$
$s = 40 m$
So, cylinder describes distance of $40 m$ before stopping.

AP EAMCET-23.08.2021

Laws of Motion

146016 Two masses $m$ and $2 m$ are hang from a frictionless. Weightless ideal pulley as shown below:

The upward acceleration of the mass $m$ is

1

\frac{g}{8}

2

\frac{g}{4}

3

\frac{g}{3}

4

\frac{g}{2}

Explanation:

C original image
And $\begin{array}{r} 2 mg - T = 21 \\ T - mg = ma \end{array}$
By adding equation (i) and (ii)
$2 mg - mg = 3 ma$
$mg = 3 ma$
$a = \frac{g}{3}$

Tripura-2020

Laws of Motion

146017 Two masses $m_{1} = 5 kg$ and $m_{2} = 4.8 kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? $(g = 9.8 m . s^{- 2})$

1

0.2 m \cdot s^{- 2}

2

9.8 m \cdot s^{- 2}

3

5.0 m . s^{- 2}

4

4.8 m \cdot s^{- 2}

Explanation:

A Given,
$m_{1} = 5 kg$
$m_{2} = 4.8 kg$
$g = 9.8 {ms}^{- 3}$
FBD for block $m_{1}$ -

$m_{1} g - T = m_{1} a$
FBD for block $m_{2} -$

$T - m_{2} g = m_{2} a$
On adding equation (i) and equation (ii), we get-
$m_{1} g - m_{2} g = m_{1} a + m_{2} a$
$g (m_{1} - m_{2}) = a (m_{1} + m_{2})$
$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
By putting the value,
$a = \frac{(5 - 4.8) \times 9.8}{5 + 4.8}$
$a = 0.2 {ms}^{- 2}$

AP EAMCET-24.09.2020

Laws of Motion

146014 In the arrangement shown in figure $a_{1}, a_{2}, a_{3}$ , and $a_{4}$ are the accelerations of masses $m_{1}, m_{2}$ , $m_{3}$ , and $m_{4}$ respectively. Which of the following relation is true for this arrangement?

1

4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0

2

a_{1} + 4 a_{2} + 3 a_{3} + a_{4} = 0

3

a_{1} + 4 a_{2} + 3 a_{3} + 2 a_{4} = 0

4

2 a_{1} + 2 a_{2} + 3 a_{3} + a_{4} = 0

Explanation:

A
Using constraints,
$Σ \vec{T} \cdot \vec{a} = 0$
$- 4 {Ta}_{1} - 2 {Ta}_{2} - {Ta}_{3} - {Ta}_{4} = 0$
$4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

JEE Main-26.06.2022

Laws of Motion

146015 A cylinder of mass $12 kg$ is sliding on plane with an initial velocity $20 {ms}^{- 1}$ . If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of

1

40 m

2

5 m

3

20 m

4

10 m

Explanation:

A Given,
$m = 12 kg$
$u = 20 {ms}^{- 1}$
$μ = 0.5$
We know that
Friction force $(f) = μ N = μ mg [∴ N = mg]$
$= 0.5 \times 12 \times 10$
$f = 60$
$∵ f = - ma$
Acceleration $(a) = - \frac{f}{m} = - \frac{60}{12} = - 5 {ms}^{- 2}$
From Newton's third law of motion,
$v^{2} = u^{2} + 2 a s$
$0 = (20)^{2} + 2 \times (- 5) \times s$
$0 = 400 - 10 s$
$s = 40 m$
So, cylinder describes distance of $40 m$ before stopping.

AP EAMCET-23.08.2021

Laws of Motion

146016 Two masses $m$ and $2 m$ are hang from a frictionless. Weightless ideal pulley as shown below:

The upward acceleration of the mass $m$ is

1

\frac{g}{8}

2

\frac{g}{4}

3

\frac{g}{3}

4

\frac{g}{2}

Explanation:

C original image
And $\begin{array}{r} 2 mg - T = 21 \\ T - mg = ma \end{array}$
By adding equation (i) and (ii)
$2 mg - mg = 3 ma$
$mg = 3 ma$
$a = \frac{g}{3}$

Tripura-2020

Laws of Motion

146017 Two masses $m_{1} = 5 kg$ and $m_{2} = 4.8 kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? $(g = 9.8 m . s^{- 2})$

1

0.2 m \cdot s^{- 2}

2

9.8 m \cdot s^{- 2}

3

5.0 m . s^{- 2}

4

4.8 m \cdot s^{- 2}

Explanation:

A Given,
$m_{1} = 5 kg$
$m_{2} = 4.8 kg$
$g = 9.8 {ms}^{- 3}$
FBD for block $m_{1}$ -

$m_{1} g - T = m_{1} a$
FBD for block $m_{2} -$

$T - m_{2} g = m_{2} a$
On adding equation (i) and equation (ii), we get-
$m_{1} g - m_{2} g = m_{1} a + m_{2} a$
$g (m_{1} - m_{2}) = a (m_{1} + m_{2})$
$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
By putting the value,
$a = \frac{(5 - 4.8) \times 9.8}{5 + 4.8}$
$a = 0.2 {ms}^{- 2}$

AP EAMCET-24.09.2020

Laws of Motion

146014 In the arrangement shown in figure $a_{1}, a_{2}, a_{3}$ , and $a_{4}$ are the accelerations of masses $m_{1}, m_{2}$ , $m_{3}$ , and $m_{4}$ respectively. Which of the following relation is true for this arrangement?

1

4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0

2

a_{1} + 4 a_{2} + 3 a_{3} + a_{4} = 0

3

a_{1} + 4 a_{2} + 3 a_{3} + 2 a_{4} = 0

4

2 a_{1} + 2 a_{2} + 3 a_{3} + a_{4} = 0

Explanation:

A
Using constraints,
$Σ \vec{T} \cdot \vec{a} = 0$
$- 4 {Ta}_{1} - 2 {Ta}_{2} - {Ta}_{3} - {Ta}_{4} = 0$
$4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

JEE Main-26.06.2022

Laws of Motion

146015 A cylinder of mass $12 kg$ is sliding on plane with an initial velocity $20 {ms}^{- 1}$ . If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of

1

40 m

2

5 m

3

20 m

4

10 m

Explanation:

A Given,
$m = 12 kg$
$u = 20 {ms}^{- 1}$
$μ = 0.5$
We know that
Friction force $(f) = μ N = μ mg [∴ N = mg]$
$= 0.5 \times 12 \times 10$
$f = 60$
$∵ f = - ma$
Acceleration $(a) = - \frac{f}{m} = - \frac{60}{12} = - 5 {ms}^{- 2}$
From Newton's third law of motion,
$v^{2} = u^{2} + 2 a s$
$0 = (20)^{2} + 2 \times (- 5) \times s$
$0 = 400 - 10 s$
$s = 40 m$
So, cylinder describes distance of $40 m$ before stopping.

AP EAMCET-23.08.2021

Laws of Motion

146016 Two masses $m$ and $2 m$ are hang from a frictionless. Weightless ideal pulley as shown below:

The upward acceleration of the mass $m$ is

1

\frac{g}{8}

2

\frac{g}{4}

3

\frac{g}{3}

4

\frac{g}{2}

Explanation:

C original image
And $\begin{array}{r} 2 mg - T = 21 \\ T - mg = ma \end{array}$
By adding equation (i) and (ii)
$2 mg - mg = 3 ma$
$mg = 3 ma$
$a = \frac{g}{3}$

Tripura-2020

Laws of Motion

146017 Two masses $m_{1} = 5 kg$ and $m_{2} = 4.8 kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? $(g = 9.8 m . s^{- 2})$

1

0.2 m \cdot s^{- 2}

2

9.8 m \cdot s^{- 2}

3

5.0 m . s^{- 2}

4

4.8 m \cdot s^{- 2}

Explanation:

A Given,
$m_{1} = 5 kg$
$m_{2} = 4.8 kg$
$g = 9.8 {ms}^{- 3}$
FBD for block $m_{1}$ -

$m_{1} g - T = m_{1} a$
FBD for block $m_{2} -$

$T - m_{2} g = m_{2} a$
On adding equation (i) and equation (ii), we get-
$m_{1} g - m_{2} g = m_{1} a + m_{2} a$
$g (m_{1} - m_{2}) = a (m_{1} + m_{2})$
$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
By putting the value,
$a = \frac{(5 - 4.8) \times 9.8}{5 + 4.8}$
$a = 0.2 {ms}^{- 2}$

AP EAMCET-24.09.2020

Laws of Motion

146014 In the arrangement shown in figure $a_{1}, a_{2}, a_{3}$ , and $a_{4}$ are the accelerations of masses $m_{1}, m_{2}$ , $m_{3}$ , and $m_{4}$ respectively. Which of the following relation is true for this arrangement?

1

4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0

2

a_{1} + 4 a_{2} + 3 a_{3} + a_{4} = 0

3

a_{1} + 4 a_{2} + 3 a_{3} + 2 a_{4} = 0

4

2 a_{1} + 2 a_{2} + 3 a_{3} + a_{4} = 0

Explanation:

A
Using constraints,
$Σ \vec{T} \cdot \vec{a} = 0$
$- 4 {Ta}_{1} - 2 {Ta}_{2} - {Ta}_{3} - {Ta}_{4} = 0$
$4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

JEE Main-26.06.2022

Laws of Motion

146015 A cylinder of mass $12 kg$ is sliding on plane with an initial velocity $20 {ms}^{- 1}$ . If the coefficient of friction between the surface and the cylinder is 0.5 , before stopping. The cylinder describes a distance of

1

40 m

2

5 m

3

20 m

4

10 m

Explanation:

A Given,
$m = 12 kg$
$u = 20 {ms}^{- 1}$
$μ = 0.5$
We know that
Friction force $(f) = μ N = μ mg [∴ N = mg]$
$= 0.5 \times 12 \times 10$
$f = 60$
$∵ f = - ma$
Acceleration $(a) = - \frac{f}{m} = - \frac{60}{12} = - 5 {ms}^{- 2}$
From Newton's third law of motion,
$v^{2} = u^{2} + 2 a s$
$0 = (20)^{2} + 2 \times (- 5) \times s$
$0 = 400 - 10 s$
$s = 40 m$
So, cylinder describes distance of $40 m$ before stopping.

AP EAMCET-23.08.2021

Laws of Motion

146016 Two masses $m$ and $2 m$ are hang from a frictionless. Weightless ideal pulley as shown below:

The upward acceleration of the mass $m$ is

1

\frac{g}{8}

2

\frac{g}{4}

3

\frac{g}{3}

4

\frac{g}{2}

Explanation:

C original image
And $\begin{array}{r} 2 mg - T = 21 \\ T - mg = ma \end{array}$
By adding equation (i) and (ii)
$2 mg - mg = 3 ma$
$mg = 3 ma$
$a = \frac{g}{3}$

Tripura-2020

Laws of Motion

146017 Two masses $m_{1} = 5 kg$ and $m_{2} = 4.8 kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses. When left free to move? $(g = 9.8 m . s^{- 2})$

1

0.2 m \cdot s^{- 2}

2

9.8 m \cdot s^{- 2}

3

5.0 m . s^{- 2}

4

4.8 m \cdot s^{- 2}

Explanation:

A Given,
$m_{1} = 5 kg$
$m_{2} = 4.8 kg$
$g = 9.8 {ms}^{- 3}$
FBD for block $m_{1}$ -

$m_{1} g - T = m_{1} a$
FBD for block $m_{2} -$

$T - m_{2} g = m_{2} a$
On adding equation (i) and equation (ii), we get-
$m_{1} g - m_{2} g = m_{1} a + m_{2} a$
$g (m_{1} - m_{2}) = a (m_{1} + m_{2})$
$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$
By putting the value,
$a = \frac{(5 - 4.8) \times 9.8}{5 + 4.8}$
$a = 0.2 {ms}^{- 2}$

AP EAMCET-24.09.2020

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