QBManager

LAWS OF MOTION (ADDITIONAL)

371953 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371954 Two blocks of masses $1 kg$ and $2 kg$ connected by a light rod and the system is slipping down rough incline angle $45^{\circ}$ with the horizontal. The frictional coefficient at both the contacts is 0.4 . If the acceleration of the system is $a \sqrt{2}$ , the value of $α$ is
(Use $g = 10 m / s^{2}$ )

1 4

2 3

3 2

4 6

Explanation:

B Given, $m_{1} = 1 kg, m_{2} = 2 kg, θ = 45^{\circ}, μ = 0.4$ According to question-

From free body diagram for $m_{1}$ block-

$N_{1} = m_{1} g \cos θ$
$m_{1} g \sin θ - T - μ N_{1} = m_{1} a$
From free body diagram for $m_{2}$ block-

$N_{2} = m_{2} g \cos θ$
$T + m_{2} g \sin θ - μ N_{2} = m_{2} a$
On adding equation (i) and (ii), we get
$(m_{1} + m_{2}) gsin θ - μ (N_{1} + N_{2}) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) g \sin θ - μ (m_{1} g \cos θ + m_{2} g \cos θ) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) gsin θ - μ g \cos θ (m_{1} + m_{2}) = (m_{1} + m_{2}) a$
$g \sin θ - μ g \cos θ = a$
$10 \times \sin 45^{\circ} - 0.4 \times 10 \times \cos 45^{\circ} = a$
$\frac{10}{\sqrt{2}} - \frac{4}{\sqrt{2}} = a$
$a = \frac{6}{\sqrt{2}}$
$a = 3 \sqrt{2}$
From question acceleration of system is $α \sqrt{2}$ . On comparing we get, $α = 3$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371955 A $10 kg$ box is pulled on a rough horizontal surface. The force $40 N$ is applied at $60^{\circ}$ angle from vertical. If co-efficient of kinetic friction is 0.25 , what will the acceleration of the moving box? (Consider $g = 10 m / s^{2}$ )

1

0.76 m / s^{2}

2

1.52 m / s^{2}

3

1.46 m / s^{2}

4

0.68 m / s^{2}

Explanation:

C Given,
Mass of box (m) $= 10 kg$
$F = 40 N$
Kinetic friction $(μ) = 0.25$
$g = 10 m / s^{2}$ and vertical angle $= 60^{\circ}$
Then, horizontal angle is $= 90^{\circ} - 60^{\circ} = 30^{\circ}$

Horizontal co-efficient of force, $F_{x} = F \cos 30^{\circ}$
$F_{x} = 40 \cos 30^{\circ}$
$F_{x} = 40 \times \frac{\sqrt{3}}{2}$
$F_{x} = 20 \sqrt{3}$
$F_{x} = 34.64 N$
Vertical co-efficient of force, $F_{y} = F \sin 30^{\circ}$
$= 40 \times 1 / 2$
$= 20 N$
Normal $= mg = 10 \times 10 = 100 N$
$F_{Normal} = 100 - 20 = 80 N$
Friction force $= μ N$
$f = 0.25 \times 80 = 20 N$
$Net force to move the object (F_{net}) = F_{x} - f$
$F_{n e t} = 34.64 - 20$
$F_{net} = 14.64$
$F_{net} = 14.64$
$∴ m \times a = 14.64 (∵ F = ma)$
$a = \frac{14.64}{10}$
$a = 1.46 m / s^{2}$

TS EAMCET 31.07.2022

LAWS OF MOTION (ADDITIONAL)

371956 A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25 , then the maximum fraction of length of the chain, that can hang over one edge of the table is-

1

20 %

2

25 %

3

35 %

4

15 %

Explanation:

A
Given, friction $(μ) = 0.25$
From fig.
$∴ f_{max} = μ \frac{M}{L} xg$
Weight of hanging chain
$W = \frac{M}{L} (L - x) g$
Equating eq $^{n}$ (i) (ii)
$μ \frac{M}{L} xg = \frac{M}{L} (L - x) g$
$μ x = (L - x)$
$0.25 x = L - x$
$1.25 x = L$
$x = \frac{1}{1.25} L = \frac{4}{5} L$
Hanging Part is L/5 which is $20 %$ of the total L
So, $A$ is correct option.

CG PET-22.05.2022

LAWS OF MOTION (ADDITIONAL)

371953 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371954 Two blocks of masses $1 kg$ and $2 kg$ connected by a light rod and the system is slipping down rough incline angle $45^{\circ}$ with the horizontal. The frictional coefficient at both the contacts is 0.4 . If the acceleration of the system is $a \sqrt{2}$ , the value of $α$ is
(Use $g = 10 m / s^{2}$ )

1 4

2 3

3 2

4 6

Explanation:

B Given, $m_{1} = 1 kg, m_{2} = 2 kg, θ = 45^{\circ}, μ = 0.4$ According to question-

From free body diagram for $m_{1}$ block-

$N_{1} = m_{1} g \cos θ$
$m_{1} g \sin θ - T - μ N_{1} = m_{1} a$
From free body diagram for $m_{2}$ block-

$N_{2} = m_{2} g \cos θ$
$T + m_{2} g \sin θ - μ N_{2} = m_{2} a$
On adding equation (i) and (ii), we get
$(m_{1} + m_{2}) gsin θ - μ (N_{1} + N_{2}) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) g \sin θ - μ (m_{1} g \cos θ + m_{2} g \cos θ) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) gsin θ - μ g \cos θ (m_{1} + m_{2}) = (m_{1} + m_{2}) a$
$g \sin θ - μ g \cos θ = a$
$10 \times \sin 45^{\circ} - 0.4 \times 10 \times \cos 45^{\circ} = a$
$\frac{10}{\sqrt{2}} - \frac{4}{\sqrt{2}} = a$
$a = \frac{6}{\sqrt{2}}$
$a = 3 \sqrt{2}$
From question acceleration of system is $α \sqrt{2}$ . On comparing we get, $α = 3$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371955 A $10 kg$ box is pulled on a rough horizontal surface. The force $40 N$ is applied at $60^{\circ}$ angle from vertical. If co-efficient of kinetic friction is 0.25 , what will the acceleration of the moving box? (Consider $g = 10 m / s^{2}$ )

1

0.76 m / s^{2}

2

1.52 m / s^{2}

3

1.46 m / s^{2}

4

0.68 m / s^{2}

Explanation:

C Given,
Mass of box (m) $= 10 kg$
$F = 40 N$
Kinetic friction $(μ) = 0.25$
$g = 10 m / s^{2}$ and vertical angle $= 60^{\circ}$
Then, horizontal angle is $= 90^{\circ} - 60^{\circ} = 30^{\circ}$

Horizontal co-efficient of force, $F_{x} = F \cos 30^{\circ}$
$F_{x} = 40 \cos 30^{\circ}$
$F_{x} = 40 \times \frac{\sqrt{3}}{2}$
$F_{x} = 20 \sqrt{3}$
$F_{x} = 34.64 N$
Vertical co-efficient of force, $F_{y} = F \sin 30^{\circ}$
$= 40 \times 1 / 2$
$= 20 N$
Normal $= mg = 10 \times 10 = 100 N$
$F_{Normal} = 100 - 20 = 80 N$
Friction force $= μ N$
$f = 0.25 \times 80 = 20 N$
$Net force to move the object (F_{net}) = F_{x} - f$
$F_{n e t} = 34.64 - 20$
$F_{net} = 14.64$
$F_{net} = 14.64$
$∴ m \times a = 14.64 (∵ F = ma)$
$a = \frac{14.64}{10}$
$a = 1.46 m / s^{2}$

TS EAMCET 31.07.2022

LAWS OF MOTION (ADDITIONAL)

371956 A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25 , then the maximum fraction of length of the chain, that can hang over one edge of the table is-

1

20 %

2

25 %

3

35 %

4

15 %

Explanation:

A
Given, friction $(μ) = 0.25$
From fig.
$∴ f_{max} = μ \frac{M}{L} xg$
Weight of hanging chain
$W = \frac{M}{L} (L - x) g$
Equating eq $^{n}$ (i) (ii)
$μ \frac{M}{L} xg = \frac{M}{L} (L - x) g$
$μ x = (L - x)$
$0.25 x = L - x$
$1.25 x = L$
$x = \frac{1}{1.25} L = \frac{4}{5} L$
Hanging Part is L/5 which is $20 %$ of the total L
So, $A$ is correct option.

CG PET-22.05.2022

LAWS OF MOTION (ADDITIONAL)

371953 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371954 Two blocks of masses $1 kg$ and $2 kg$ connected by a light rod and the system is slipping down rough incline angle $45^{\circ}$ with the horizontal. The frictional coefficient at both the contacts is 0.4 . If the acceleration of the system is $a \sqrt{2}$ , the value of $α$ is
(Use $g = 10 m / s^{2}$ )

1 4

2 3

3 2

4 6

Explanation:

B Given, $m_{1} = 1 kg, m_{2} = 2 kg, θ = 45^{\circ}, μ = 0.4$ According to question-

From free body diagram for $m_{1}$ block-

$N_{1} = m_{1} g \cos θ$
$m_{1} g \sin θ - T - μ N_{1} = m_{1} a$
From free body diagram for $m_{2}$ block-

$N_{2} = m_{2} g \cos θ$
$T + m_{2} g \sin θ - μ N_{2} = m_{2} a$
On adding equation (i) and (ii), we get
$(m_{1} + m_{2}) gsin θ - μ (N_{1} + N_{2}) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) g \sin θ - μ (m_{1} g \cos θ + m_{2} g \cos θ) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) gsin θ - μ g \cos θ (m_{1} + m_{2}) = (m_{1} + m_{2}) a$
$g \sin θ - μ g \cos θ = a$
$10 \times \sin 45^{\circ} - 0.4 \times 10 \times \cos 45^{\circ} = a$
$\frac{10}{\sqrt{2}} - \frac{4}{\sqrt{2}} = a$
$a = \frac{6}{\sqrt{2}}$
$a = 3 \sqrt{2}$
From question acceleration of system is $α \sqrt{2}$ . On comparing we get, $α = 3$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371955 A $10 kg$ box is pulled on a rough horizontal surface. The force $40 N$ is applied at $60^{\circ}$ angle from vertical. If co-efficient of kinetic friction is 0.25 , what will the acceleration of the moving box? (Consider $g = 10 m / s^{2}$ )

1

0.76 m / s^{2}

2

1.52 m / s^{2}

3

1.46 m / s^{2}

4

0.68 m / s^{2}

Explanation:

C Given,
Mass of box (m) $= 10 kg$
$F = 40 N$
Kinetic friction $(μ) = 0.25$
$g = 10 m / s^{2}$ and vertical angle $= 60^{\circ}$
Then, horizontal angle is $= 90^{\circ} - 60^{\circ} = 30^{\circ}$

Horizontal co-efficient of force, $F_{x} = F \cos 30^{\circ}$
$F_{x} = 40 \cos 30^{\circ}$
$F_{x} = 40 \times \frac{\sqrt{3}}{2}$
$F_{x} = 20 \sqrt{3}$
$F_{x} = 34.64 N$
Vertical co-efficient of force, $F_{y} = F \sin 30^{\circ}$
$= 40 \times 1 / 2$
$= 20 N$
Normal $= mg = 10 \times 10 = 100 N$
$F_{Normal} = 100 - 20 = 80 N$
Friction force $= μ N$
$f = 0.25 \times 80 = 20 N$
$Net force to move the object (F_{net}) = F_{x} - f$
$F_{n e t} = 34.64 - 20$
$F_{net} = 14.64$
$F_{net} = 14.64$
$∴ m \times a = 14.64 (∵ F = ma)$
$a = \frac{14.64}{10}$
$a = 1.46 m / s^{2}$

TS EAMCET 31.07.2022

LAWS OF MOTION (ADDITIONAL)

371956 A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25 , then the maximum fraction of length of the chain, that can hang over one edge of the table is-

1

20 %

2

25 %

3

35 %

4

15 %

Explanation:

A
Given, friction $(μ) = 0.25$
From fig.
$∴ f_{max} = μ \frac{M}{L} xg$
Weight of hanging chain
$W = \frac{M}{L} (L - x) g$
Equating eq $^{n}$ (i) (ii)
$μ \frac{M}{L} xg = \frac{M}{L} (L - x) g$
$μ x = (L - x)$
$0.25 x = L - x$
$1.25 x = L$
$x = \frac{1}{1.25} L = \frac{4}{5} L$
Hanging Part is L/5 which is $20 %$ of the total L
So, $A$ is correct option.

CG PET-22.05.2022

LAWS OF MOTION (ADDITIONAL)

371953 A block is placed on a parabolic shape ramp given by equation $y = \frac{x^{2}}{20}$ . If the coefficient of static friction $(μ_{s})$ is 0.5 . then what is the maximum height above the ground at which the block can be placed without slipping?

1

2.5 m

2

1.25 m

3

0.5 m

4

0.25 m

Explanation:

B Limiting friction force (f) $= μ N = μ_{s} mg \cos θ$

In equilibrium, $mg \sin θ = μ_{s} mg \cos θ$
$\frac{\sin θ}{\cos θ} = μ_{s}$
$\tan θ = μ_{s}$
$\tan θ = 0.5$
Given, $y = \frac{x^{2}}{20}$
$\frac{dy}{dx} = \frac{2 x}{20} = \frac{x}{10}$
$\tan θ = \frac{x}{10}$
Putting value of $\tan θ$ in equation (ii), we get
$0.5 = \frac{x}{10}$
$x = 5$
Putting value of $x$ in equation (i), we get
$h = y_{max} = \frac{x^{2}}{20} = \frac{(5)^{2}}{20} = 1.25 m$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371954 Two blocks of masses $1 kg$ and $2 kg$ connected by a light rod and the system is slipping down rough incline angle $45^{\circ}$ with the horizontal. The frictional coefficient at both the contacts is 0.4 . If the acceleration of the system is $a \sqrt{2}$ , the value of $α$ is
(Use $g = 10 m / s^{2}$ )

1 4

2 3

3 2

4 6

Explanation:

B Given, $m_{1} = 1 kg, m_{2} = 2 kg, θ = 45^{\circ}, μ = 0.4$ According to question-

From free body diagram for $m_{1}$ block-

$N_{1} = m_{1} g \cos θ$
$m_{1} g \sin θ - T - μ N_{1} = m_{1} a$
From free body diagram for $m_{2}$ block-

$N_{2} = m_{2} g \cos θ$
$T + m_{2} g \sin θ - μ N_{2} = m_{2} a$
On adding equation (i) and (ii), we get
$(m_{1} + m_{2}) gsin θ - μ (N_{1} + N_{2}) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) g \sin θ - μ (m_{1} g \cos θ + m_{2} g \cos θ) = (m_{1} + m_{2}) a$
$(m_{1} + m_{2}) gsin θ - μ g \cos θ (m_{1} + m_{2}) = (m_{1} + m_{2}) a$
$g \sin θ - μ g \cos θ = a$
$10 \times \sin 45^{\circ} - 0.4 \times 10 \times \cos 45^{\circ} = a$
$\frac{10}{\sqrt{2}} - \frac{4}{\sqrt{2}} = a$
$a = \frac{6}{\sqrt{2}}$
$a = 3 \sqrt{2}$
From question acceleration of system is $α \sqrt{2}$ . On comparing we get, $α = 3$

TS EAMCET 18.07.2022

LAWS OF MOTION (ADDITIONAL)

371955 A $10 kg$ box is pulled on a rough horizontal surface. The force $40 N$ is applied at $60^{\circ}$ angle from vertical. If co-efficient of kinetic friction is 0.25 , what will the acceleration of the moving box? (Consider $g = 10 m / s^{2}$ )

1

0.76 m / s^{2}

2

1.52 m / s^{2}

3

1.46 m / s^{2}

4

0.68 m / s^{2}

Explanation:

C Given,
Mass of box (m) $= 10 kg$
$F = 40 N$
Kinetic friction $(μ) = 0.25$
$g = 10 m / s^{2}$ and vertical angle $= 60^{\circ}$
Then, horizontal angle is $= 90^{\circ} - 60^{\circ} = 30^{\circ}$

Horizontal co-efficient of force, $F_{x} = F \cos 30^{\circ}$
$F_{x} = 40 \cos 30^{\circ}$
$F_{x} = 40 \times \frac{\sqrt{3}}{2}$
$F_{x} = 20 \sqrt{3}$
$F_{x} = 34.64 N$
Vertical co-efficient of force, $F_{y} = F \sin 30^{\circ}$
$= 40 \times 1 / 2$
$= 20 N$
Normal $= mg = 10 \times 10 = 100 N$
$F_{Normal} = 100 - 20 = 80 N$
Friction force $= μ N$
$f = 0.25 \times 80 = 20 N$
$Net force to move the object (F_{net}) = F_{x} - f$
$F_{n e t} = 34.64 - 20$
$F_{net} = 14.64$
$F_{net} = 14.64$
$∴ m \times a = 14.64 (∵ F = ma)$
$a = \frac{14.64}{10}$
$a = 1.46 m / s^{2}$

TS EAMCET 31.07.2022

LAWS OF MOTION (ADDITIONAL)

371956 A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25 , then the maximum fraction of length of the chain, that can hang over one edge of the table is-

1

20 %

2

25 %

3

35 %

4

15 %

Explanation:

A
Given, friction $(μ) = 0.25$
From fig.
$∴ f_{max} = μ \frac{M}{L} xg$
Weight of hanging chain
$W = \frac{M}{L} (L - x) g$
Equating eq $^{n}$ (i) (ii)
$μ \frac{M}{L} xg = \frac{M}{L} (L - x) g$
$μ x = (L - x)$
$0.25 x = L - x$
$1.25 x = L$
$x = \frac{1}{1.25} L = \frac{4}{5} L$
Hanging Part is L/5 which is $20 %$ of the total L
So, $A$ is correct option.

CG PET-22.05.2022