QBManager

Laws of Motion

146178 A child is on a merry-go-round, standing at a distance of $2 m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-goround be rotated before the child slips? (Take, $g = 10 m / s^{2}$ )

1

0.5 rad / s

2

1 rad / s

3

2 rad / s

4

4 rad / s

Explanation:

C
Given, $μ = 0.8, r = 2 m$
Centripetal force $F_{c} = \frac{{mv}^{2}}{r} = mr ω^{2}$
Friction force -
$f = N μ = μ mg$
At critical speed, maximum static friction must be equal to centrifugal force on child.
$μ N = mr ω^{2}$
$0.8 \times m \times 10 = m \cdot (2) \cdot ω^{2}$
$8 = 2 ω^{2}$
$ω^{2} = \frac{8}{2}$
$ω^{2} = 4$
$ω = \sqrt{4} = 2 rad / s$

TS- EAMCET-07.05.2018

Laws of Motion

146179 A movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is 0.57 and for steel on brass is 0.44. Assuming that the entire weight comes into the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$ ) is

1 13

2 101

3 440

4 570

Explanation:

B
Given,
$w = 100 N, N = 100 N$
Friction force between steel-
$f_{s - s} = μ_{s} \times N = 0.57 \times 100 = 57 N$

Friction force between steel and brass -

$F \geq f_{s - s} + f_{s - b}$
$F \geq 57 + 44 \Rightarrow F \geq 101 N$
$F = 101 N$ .

TS- EAMCET-04.05.2018

Laws of Motion

146180 A block of mass $5 kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

1

12 N

2

22 N

3

32 N

4

42 N

Explanation:

B Given, $m = 5 kg, μ = 0.1, a = 3 m / s^{2}$

In vertical direction,
$N = mg - F \sin 30^{\circ}$
Net force on block in horizontal direction,
$ma = F \cos 30^{\circ} - f_{s}$
$ma = F \cos 30^{\circ} - μ N$
$= F \cos 30^{\circ} - μ (mg - F \sin 30^{\circ})$
$5 \times 3 = F \times \frac{\sqrt{3}}{2} - 0.1 (5 \times 10 - F \times \frac{1}{2})$
$15 = \frac{\sqrt{3} F}{2} - 5 + \frac{0.1 F}{2}$
$30 = \sqrt{3} F - 10 + 0.1 F$
$1.8 F = 40 \Rightarrow F = \frac{40}{1.8} = 22.22 N \approx 22 N$

TS- EAMCET-05.05.2018

Laws of Motion

146181 Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . coefficient of static friction between the sand layers is $μ$ . Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

D Let, $h =$ height of sand cone $θ =$ semi - vertical angle of cone $R =$ base radius

Consider a sand particle at point $P$ In equilibrium condition $N = mgsin θ$ We know, Friction force $(f) = μ N$ $f = μ mg \sin θ \dots . . (i)$
From figure
$mg \cos θ = f$
Putting the value of ' $f$ ' in equation (ii)
$mg \cos θ = μ mg \sin θ$
$\frac{\sin θ}{\cos θ} = \frac{1}{μ}$
$\tan θ = \frac{1}{μ}$
From triangle $AOC$
$\tan θ = \frac{O C}{A O} = \frac{R}{h}$
$\tan θ = \frac{R}{h} = \frac{1}{μ} \Rightarrow h = μ R$
Therefore maximum volume of sand cone
$V_{max} = \frac{1}{3} π r^{2} h$
Then, $V_{max} = \frac{1}{3} π R^{2} (μ R) = \frac{μ π R^{3}}{3}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146182 A block of mass $2 kg$ is being pushed against a wall by a force $F = 90 N$ as shown in the figure. If the coefficient of friction is 0.25 . Then the magnitude of acceleration of the block is (Take,
$g = 10 {m s}^{- 2}) (\sin 37^{\circ} = \frac{3}{5})$
original image

1

16 {ms}^{- 2}

2

8 {ms}^{- 2}

3

38 {ms}^{- 2}

4

54 {ms}^{- 2}

Explanation:

B Given, Coefficient of friction $(μ) = 0.25$ mass $(m) = 2 kg$ , applied force $(F) = 90 N, \sin 37^{\circ} = \frac{3}{5}$ and $\cos 37^{\circ} = \frac{4}{5}$

Block move upward
Resolving the forces along $x$ and $y$ direction -
Net horizontal force, $Σ F_{x} = 0$
$N = F \cos (37^{\circ})$
$N = 90 \times \frac{4}{5} = 54 N$
Friction force $(f_{r}) = μ N = 0.25 \times 90 \times \frac{4}{5} {N = F \cos 37^{\circ}}$
$f_{r} = 18 N$
Net vertical force, $Σ F_{y} = 0$
$F \sin 37^{\circ} - mg - f_{r} = ma$
$90 \times \frac{3}{5} - 2 \times 10 - 18 = 2 a$
$54 - 38 = 2 a$
$a = 8 m / \sec^{2}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146178 A child is on a merry-go-round, standing at a distance of $2 m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-goround be rotated before the child slips? (Take, $g = 10 m / s^{2}$ )

1

0.5 rad / s

2

1 rad / s

3

2 rad / s

4

4 rad / s

Explanation:

C
Given, $μ = 0.8, r = 2 m$
Centripetal force $F_{c} = \frac{{mv}^{2}}{r} = mr ω^{2}$
Friction force -
$f = N μ = μ mg$
At critical speed, maximum static friction must be equal to centrifugal force on child.
$μ N = mr ω^{2}$
$0.8 \times m \times 10 = m \cdot (2) \cdot ω^{2}$
$8 = 2 ω^{2}$
$ω^{2} = \frac{8}{2}$
$ω^{2} = 4$
$ω = \sqrt{4} = 2 rad / s$

TS- EAMCET-07.05.2018

Laws of Motion

146179 A movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is 0.57 and for steel on brass is 0.44. Assuming that the entire weight comes into the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$ ) is

1 13

2 101

3 440

4 570

Explanation:

B
Given,
$w = 100 N, N = 100 N$
Friction force between steel-
$f_{s - s} = μ_{s} \times N = 0.57 \times 100 = 57 N$

Friction force between steel and brass -

$F \geq f_{s - s} + f_{s - b}$
$F \geq 57 + 44 \Rightarrow F \geq 101 N$
$F = 101 N$ .

TS- EAMCET-04.05.2018

Laws of Motion

146180 A block of mass $5 kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

1

12 N

2

22 N

3

32 N

4

42 N

Explanation:

B Given, $m = 5 kg, μ = 0.1, a = 3 m / s^{2}$

In vertical direction,
$N = mg - F \sin 30^{\circ}$
Net force on block in horizontal direction,
$ma = F \cos 30^{\circ} - f_{s}$
$ma = F \cos 30^{\circ} - μ N$
$= F \cos 30^{\circ} - μ (mg - F \sin 30^{\circ})$
$5 \times 3 = F \times \frac{\sqrt{3}}{2} - 0.1 (5 \times 10 - F \times \frac{1}{2})$
$15 = \frac{\sqrt{3} F}{2} - 5 + \frac{0.1 F}{2}$
$30 = \sqrt{3} F - 10 + 0.1 F$
$1.8 F = 40 \Rightarrow F = \frac{40}{1.8} = 22.22 N \approx 22 N$

TS- EAMCET-05.05.2018

Laws of Motion

146181 Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . coefficient of static friction between the sand layers is $μ$ . Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

D Let, $h =$ height of sand cone $θ =$ semi - vertical angle of cone $R =$ base radius

Consider a sand particle at point $P$ In equilibrium condition $N = mgsin θ$ We know, Friction force $(f) = μ N$ $f = μ mg \sin θ \dots . . (i)$
From figure
$mg \cos θ = f$
Putting the value of ' $f$ ' in equation (ii)
$mg \cos θ = μ mg \sin θ$
$\frac{\sin θ}{\cos θ} = \frac{1}{μ}$
$\tan θ = \frac{1}{μ}$
From triangle $AOC$
$\tan θ = \frac{O C}{A O} = \frac{R}{h}$
$\tan θ = \frac{R}{h} = \frac{1}{μ} \Rightarrow h = μ R$
Therefore maximum volume of sand cone
$V_{max} = \frac{1}{3} π r^{2} h$
Then, $V_{max} = \frac{1}{3} π R^{2} (μ R) = \frac{μ π R^{3}}{3}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146182 A block of mass $2 kg$ is being pushed against a wall by a force $F = 90 N$ as shown in the figure. If the coefficient of friction is 0.25 . Then the magnitude of acceleration of the block is (Take,
$g = 10 {m s}^{- 2}) (\sin 37^{\circ} = \frac{3}{5})$
original image

1

16 {ms}^{- 2}

2

8 {ms}^{- 2}

3

38 {ms}^{- 2}

4

54 {ms}^{- 2}

Explanation:

B Given, Coefficient of friction $(μ) = 0.25$ mass $(m) = 2 kg$ , applied force $(F) = 90 N, \sin 37^{\circ} = \frac{3}{5}$ and $\cos 37^{\circ} = \frac{4}{5}$

Block move upward
Resolving the forces along $x$ and $y$ direction -
Net horizontal force, $Σ F_{x} = 0$
$N = F \cos (37^{\circ})$
$N = 90 \times \frac{4}{5} = 54 N$
Friction force $(f_{r}) = μ N = 0.25 \times 90 \times \frac{4}{5} {N = F \cos 37^{\circ}}$
$f_{r} = 18 N$
Net vertical force, $Σ F_{y} = 0$
$F \sin 37^{\circ} - mg - f_{r} = ma$
$90 \times \frac{3}{5} - 2 \times 10 - 18 = 2 a$
$54 - 38 = 2 a$
$a = 8 m / \sec^{2}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146178 A child is on a merry-go-round, standing at a distance of $2 m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-goround be rotated before the child slips? (Take, $g = 10 m / s^{2}$ )

1

0.5 rad / s

2

1 rad / s

3

2 rad / s

4

4 rad / s

Explanation:

C
Given, $μ = 0.8, r = 2 m$
Centripetal force $F_{c} = \frac{{mv}^{2}}{r} = mr ω^{2}$
Friction force -
$f = N μ = μ mg$
At critical speed, maximum static friction must be equal to centrifugal force on child.
$μ N = mr ω^{2}$
$0.8 \times m \times 10 = m \cdot (2) \cdot ω^{2}$
$8 = 2 ω^{2}$
$ω^{2} = \frac{8}{2}$
$ω^{2} = 4$
$ω = \sqrt{4} = 2 rad / s$

TS- EAMCET-07.05.2018

Laws of Motion

146179 A movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is 0.57 and for steel on brass is 0.44. Assuming that the entire weight comes into the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$ ) is

1 13

2 101

3 440

4 570

Explanation:

B
Given,
$w = 100 N, N = 100 N$
Friction force between steel-
$f_{s - s} = μ_{s} \times N = 0.57 \times 100 = 57 N$

Friction force between steel and brass -

$F \geq f_{s - s} + f_{s - b}$
$F \geq 57 + 44 \Rightarrow F \geq 101 N$
$F = 101 N$ .

TS- EAMCET-04.05.2018

Laws of Motion

146180 A block of mass $5 kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

1

12 N

2

22 N

3

32 N

4

42 N

Explanation:

B Given, $m = 5 kg, μ = 0.1, a = 3 m / s^{2}$

In vertical direction,
$N = mg - F \sin 30^{\circ}$
Net force on block in horizontal direction,
$ma = F \cos 30^{\circ} - f_{s}$
$ma = F \cos 30^{\circ} - μ N$
$= F \cos 30^{\circ} - μ (mg - F \sin 30^{\circ})$
$5 \times 3 = F \times \frac{\sqrt{3}}{2} - 0.1 (5 \times 10 - F \times \frac{1}{2})$
$15 = \frac{\sqrt{3} F}{2} - 5 + \frac{0.1 F}{2}$
$30 = \sqrt{3} F - 10 + 0.1 F$
$1.8 F = 40 \Rightarrow F = \frac{40}{1.8} = 22.22 N \approx 22 N$

TS- EAMCET-05.05.2018

Laws of Motion

146181 Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . coefficient of static friction between the sand layers is $μ$ . Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

D Let, $h =$ height of sand cone $θ =$ semi - vertical angle of cone $R =$ base radius

Consider a sand particle at point $P$ In equilibrium condition $N = mgsin θ$ We know, Friction force $(f) = μ N$ $f = μ mg \sin θ \dots . . (i)$
From figure
$mg \cos θ = f$
Putting the value of ' $f$ ' in equation (ii)
$mg \cos θ = μ mg \sin θ$
$\frac{\sin θ}{\cos θ} = \frac{1}{μ}$
$\tan θ = \frac{1}{μ}$
From triangle $AOC$
$\tan θ = \frac{O C}{A O} = \frac{R}{h}$
$\tan θ = \frac{R}{h} = \frac{1}{μ} \Rightarrow h = μ R$
Therefore maximum volume of sand cone
$V_{max} = \frac{1}{3} π r^{2} h$
Then, $V_{max} = \frac{1}{3} π R^{2} (μ R) = \frac{μ π R^{3}}{3}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146182 A block of mass $2 kg$ is being pushed against a wall by a force $F = 90 N$ as shown in the figure. If the coefficient of friction is 0.25 . Then the magnitude of acceleration of the block is (Take,
$g = 10 {m s}^{- 2}) (\sin 37^{\circ} = \frac{3}{5})$
original image

1

16 {ms}^{- 2}

2

8 {ms}^{- 2}

3

38 {ms}^{- 2}

4

54 {ms}^{- 2}

Explanation:

B Given, Coefficient of friction $(μ) = 0.25$ mass $(m) = 2 kg$ , applied force $(F) = 90 N, \sin 37^{\circ} = \frac{3}{5}$ and $\cos 37^{\circ} = \frac{4}{5}$

Block move upward
Resolving the forces along $x$ and $y$ direction -
Net horizontal force, $Σ F_{x} = 0$
$N = F \cos (37^{\circ})$
$N = 90 \times \frac{4}{5} = 54 N$
Friction force $(f_{r}) = μ N = 0.25 \times 90 \times \frac{4}{5} {N = F \cos 37^{\circ}}$
$f_{r} = 18 N$
Net vertical force, $Σ F_{y} = 0$
$F \sin 37^{\circ} - mg - f_{r} = ma$
$90 \times \frac{3}{5} - 2 \times 10 - 18 = 2 a$
$54 - 38 = 2 a$
$a = 8 m / \sec^{2}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146178 A child is on a merry-go-round, standing at a distance of $2 m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-goround be rotated before the child slips? (Take, $g = 10 m / s^{2}$ )

1

0.5 rad / s

2

1 rad / s

3

2 rad / s

4

4 rad / s

Explanation:

C
Given, $μ = 0.8, r = 2 m$
Centripetal force $F_{c} = \frac{{mv}^{2}}{r} = mr ω^{2}$
Friction force -
$f = N μ = μ mg$
At critical speed, maximum static friction must be equal to centrifugal force on child.
$μ N = mr ω^{2}$
$0.8 \times m \times 10 = m \cdot (2) \cdot ω^{2}$
$8 = 2 ω^{2}$
$ω^{2} = \frac{8}{2}$
$ω^{2} = 4$
$ω = \sqrt{4} = 2 rad / s$

TS- EAMCET-07.05.2018

Laws of Motion

146179 A movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is 0.57 and for steel on brass is 0.44. Assuming that the entire weight comes into the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$ ) is

1 13

2 101

3 440

4 570

Explanation:

B
Given,
$w = 100 N, N = 100 N$
Friction force between steel-
$f_{s - s} = μ_{s} \times N = 0.57 \times 100 = 57 N$

Friction force between steel and brass -

$F \geq f_{s - s} + f_{s - b}$
$F \geq 57 + 44 \Rightarrow F \geq 101 N$
$F = 101 N$ .

TS- EAMCET-04.05.2018

Laws of Motion

146180 A block of mass $5 kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

1

12 N

2

22 N

3

32 N

4

42 N

Explanation:

B Given, $m = 5 kg, μ = 0.1, a = 3 m / s^{2}$

In vertical direction,
$N = mg - F \sin 30^{\circ}$
Net force on block in horizontal direction,
$ma = F \cos 30^{\circ} - f_{s}$
$ma = F \cos 30^{\circ} - μ N$
$= F \cos 30^{\circ} - μ (mg - F \sin 30^{\circ})$
$5 \times 3 = F \times \frac{\sqrt{3}}{2} - 0.1 (5 \times 10 - F \times \frac{1}{2})$
$15 = \frac{\sqrt{3} F}{2} - 5 + \frac{0.1 F}{2}$
$30 = \sqrt{3} F - 10 + 0.1 F$
$1.8 F = 40 \Rightarrow F = \frac{40}{1.8} = 22.22 N \approx 22 N$

TS- EAMCET-05.05.2018

Laws of Motion

146181 Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . coefficient of static friction between the sand layers is $μ$ . Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

D Let, $h =$ height of sand cone $θ =$ semi - vertical angle of cone $R =$ base radius

Consider a sand particle at point $P$ In equilibrium condition $N = mgsin θ$ We know, Friction force $(f) = μ N$ $f = μ mg \sin θ \dots . . (i)$
From figure
$mg \cos θ = f$
Putting the value of ' $f$ ' in equation (ii)
$mg \cos θ = μ mg \sin θ$
$\frac{\sin θ}{\cos θ} = \frac{1}{μ}$
$\tan θ = \frac{1}{μ}$
From triangle $AOC$
$\tan θ = \frac{O C}{A O} = \frac{R}{h}$
$\tan θ = \frac{R}{h} = \frac{1}{μ} \Rightarrow h = μ R$
Therefore maximum volume of sand cone
$V_{max} = \frac{1}{3} π r^{2} h$
Then, $V_{max} = \frac{1}{3} π R^{2} (μ R) = \frac{μ π R^{3}}{3}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146182 A block of mass $2 kg$ is being pushed against a wall by a force $F = 90 N$ as shown in the figure. If the coefficient of friction is 0.25 . Then the magnitude of acceleration of the block is (Take,
$g = 10 {m s}^{- 2}) (\sin 37^{\circ} = \frac{3}{5})$
original image

1

16 {ms}^{- 2}

2

8 {ms}^{- 2}

3

38 {ms}^{- 2}

4

54 {ms}^{- 2}

Explanation:

B Given, Coefficient of friction $(μ) = 0.25$ mass $(m) = 2 kg$ , applied force $(F) = 90 N, \sin 37^{\circ} = \frac{3}{5}$ and $\cos 37^{\circ} = \frac{4}{5}$

Block move upward
Resolving the forces along $x$ and $y$ direction -
Net horizontal force, $Σ F_{x} = 0$
$N = F \cos (37^{\circ})$
$N = 90 \times \frac{4}{5} = 54 N$
Friction force $(f_{r}) = μ N = 0.25 \times 90 \times \frac{4}{5} {N = F \cos 37^{\circ}}$
$f_{r} = 18 N$
Net vertical force, $Σ F_{y} = 0$
$F \sin 37^{\circ} - mg - f_{r} = ma$
$90 \times \frac{3}{5} - 2 \times 10 - 18 = 2 a$
$54 - 38 = 2 a$
$a = 8 m / \sec^{2}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146178 A child is on a merry-go-round, standing at a distance of $2 m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-goround be rotated before the child slips? (Take, $g = 10 m / s^{2}$ )

1

0.5 rad / s

2

1 rad / s

3

2 rad / s

4

4 rad / s

Explanation:

C
Given, $μ = 0.8, r = 2 m$
Centripetal force $F_{c} = \frac{{mv}^{2}}{r} = mr ω^{2}$
Friction force -
$f = N μ = μ mg$
At critical speed, maximum static friction must be equal to centrifugal force on child.
$μ N = mr ω^{2}$
$0.8 \times m \times 10 = m \cdot (2) \cdot ω^{2}$
$8 = 2 ω^{2}$
$ω^{2} = \frac{8}{2}$
$ω^{2} = 4$
$ω = \sqrt{4} = 2 rad / s$

TS- EAMCET-07.05.2018

Laws of Motion

146179 A movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is 0.57 and for steel on brass is 0.44. Assuming that the entire weight comes into the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$ ) is

1 13

2 101

3 440

4 570

Explanation:

B
Given,
$w = 100 N, N = 100 N$
Friction force between steel-
$f_{s - s} = μ_{s} \times N = 0.57 \times 100 = 57 N$

Friction force between steel and brass -

$F \geq f_{s - s} + f_{s - b}$
$F \geq 57 + 44 \Rightarrow F \geq 101 N$
$F = 101 N$ .

TS- EAMCET-04.05.2018

Laws of Motion

146180 A block of mass $5 kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

1

12 N

2

22 N

3

32 N

4

42 N

Explanation:

B Given, $m = 5 kg, μ = 0.1, a = 3 m / s^{2}$

In vertical direction,
$N = mg - F \sin 30^{\circ}$
Net force on block in horizontal direction,
$ma = F \cos 30^{\circ} - f_{s}$
$ma = F \cos 30^{\circ} - μ N$
$= F \cos 30^{\circ} - μ (mg - F \sin 30^{\circ})$
$5 \times 3 = F \times \frac{\sqrt{3}}{2} - 0.1 (5 \times 10 - F \times \frac{1}{2})$
$15 = \frac{\sqrt{3} F}{2} - 5 + \frac{0.1 F}{2}$
$30 = \sqrt{3} F - 10 + 0.1 F$
$1.8 F = 40 \Rightarrow F = \frac{40}{1.8} = 22.22 N \approx 22 N$

TS- EAMCET-05.05.2018

Laws of Motion

146181 Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$ . coefficient of static friction between the sand layers is $μ$ . Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

1

\frac{μ R^{3}}{3 π}

2

\frac{μ R^{3}}{3}

3

\frac{π R^{3}}{3 μ}

4

\frac{μ π R^{3}}{3}

Explanation:

D Let, $h =$ height of sand cone $θ =$ semi - vertical angle of cone $R =$ base radius

Consider a sand particle at point $P$ In equilibrium condition $N = mgsin θ$ We know, Friction force $(f) = μ N$ $f = μ mg \sin θ \dots . . (i)$
From figure
$mg \cos θ = f$
Putting the value of ' $f$ ' in equation (ii)
$mg \cos θ = μ mg \sin θ$
$\frac{\sin θ}{\cos θ} = \frac{1}{μ}$
$\tan θ = \frac{1}{μ}$
From triangle $AOC$
$\tan θ = \frac{O C}{A O} = \frac{R}{h}$
$\tan θ = \frac{R}{h} = \frac{1}{μ} \Rightarrow h = μ R$
Therefore maximum volume of sand cone
$V_{max} = \frac{1}{3} π r^{2} h$
Then, $V_{max} = \frac{1}{3} π R^{2} (μ R) = \frac{μ π R^{3}}{3}$

AP EAMCET (22.04.2018) Shift-1

Laws of Motion

146182 A block of mass $2 kg$ is being pushed against a wall by a force $F = 90 N$ as shown in the figure. If the coefficient of friction is 0.25 . Then the magnitude of acceleration of the block is (Take,
$g = 10 {m s}^{- 2}) (\sin 37^{\circ} = \frac{3}{5})$
original image

1

16 {ms}^{- 2}

2

8 {ms}^{- 2}

3

38 {ms}^{- 2}

4

54 {ms}^{- 2}

Explanation:

B Given, Coefficient of friction $(μ) = 0.25$ mass $(m) = 2 kg$ , applied force $(F) = 90 N, \sin 37^{\circ} = \frac{3}{5}$ and $\cos 37^{\circ} = \frac{4}{5}$

Block move upward
Resolving the forces along $x$ and $y$ direction -
Net horizontal force, $Σ F_{x} = 0$
$N = F \cos (37^{\circ})$
$N = 90 \times \frac{4}{5} = 54 N$
Friction force $(f_{r}) = μ N = 0.25 \times 90 \times \frac{4}{5} {N = F \cos 37^{\circ}}$
$f_{r} = 18 N$
Net vertical force, $Σ F_{y} = 0$
$F \sin 37^{\circ} - mg - f_{r} = ma$
$90 \times \frac{3}{5} - 2 \times 10 - 18 = 2 a$
$54 - 38 = 2 a$
$a = 8 m / \sec^{2}$

AP EAMCET (22.04.2018) Shift-1

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: