QBManager

LAWS OF MOTION (ADDITIONAL)

372171 A block of mass $m$ is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is $μ$ . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction

1 zero

2

2 μ mg

3

\frac{1}{2} μ mg

4

μ mg

Explanation:

A When lift freely fall, $N = 0$
We know that, $f = μ_{s} N$
$f = μ_{s} \times 0$
Force of friction $f = 0$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372172 The coefficient of friction between a hemispherical bowl and an insect is $\sqrt{0.44}$ and the radius of the bowl is $0.6 m$ . The maximum height to which an insect can crawl in the bowl will be

1

0.4 m

2

0.2 m

3

0.3 m

4

0.1 m

Explanation:

D $R = 0.6 cm$
$μ = \sqrt{0.44}$

From figure : $R = mg \cos θ$
$f = mg \sin θ$
$\Rightarrow \frac{f}{R} = \frac{mg \sin θ}{mg \cos θ} = \tan θ$
Also, $f = μ R \Rightarrow \frac{f}{R} = μ$
Hence, $\tan θ = μ$
$h = R - R \cos θ = R [1 - \cos θ]$
$= R [1 - \frac{1}{\sqrt{1 + μ^{2}}}]$
$\Rightarrow h = 0.6 [1 - \frac{1}{\sqrt{1 + 0.44}}] = 0.6 [\frac{1}{6}]$

$\tan θ = μ$
$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}$
$h = 0.1 cm$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372173 The rear side of truck is open and a box of mass $50 kg$ is placed at $3.2 m$ away from the open end. The coefficient of friction between the box. and the surface is 0.2 . If the truck starts from rest and moves on a straight road with acceleration $2.1 {ms}^{- 2}$ , then the box falls off the truck after a time $($ take $g = 10 {ms}^{- 2})$

1

4 s

2

\frac{8}{\sqrt{41}} s

3

8 s

4

\sqrt{8} s

Explanation:

C Acceleration due to pseudo force, $a_{p} = 2.1$ $m / \sec^{2}$
Net acceleration, $a = a_{p} - μ g$
Acceleration due to slipping between box and surface.
$a = 2.1 - (0.2 \times 10) = 0.1$
$t = \sqrt{\frac{2 s}{a}}$
Where, $a = 0.1$ and $s = 3.2$
$t = \sqrt{\frac{2 (3.2)}{0.1}} = \sqrt{64}$
$t = 8 \sec$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372174 A car is traveling along a curved road of radius $r$ . If the coefficient of friction between the tyres and the road is $μ$ , the car will skid if its speed exceeds

1

2 \sqrt{μ rg}

2

3 \sqrt{μ rg}

3

\sqrt{2 μ rg}

4

\sqrt{μ rg}

Explanation:

D
Let, Mass of car $= m$
Radius of curved road $= r$
For curved road-
Centripetal force $=$ Psudo force
$F_{c} = F_{r}$
$\frac{{mv}^{2}}{r} = μ mg$
Speed $v = \sqrt{μ r g}$

AP EMCET(Medical)-2010

LAWS OF MOTION (ADDITIONAL)

372175 A body of mass $10 kg$ is lying on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal and the coefficient of friction is 0.5 . The minimum force required to pull the body up the plane is

1

914 N

2

91.4 N

3

9.14 N

4

0.914 N

Explanation:

B
We know that,
$f_{k} = μ N$
$= μ mg \cos θ$
According to question-
$F = f_{k} + mg \sin θ$
$= μ mg \cos θ + mg \sin θ$
$= μ mg \cos 30^{\circ} + mg \sin 30^{\circ}$
$= 5 g \times \frac{\sqrt{3}}{2} + 10 g \times \frac{1}{2}$
$= 5 g + 5 g \times \frac{\sqrt{3}}{2}$
$= 5 g (1 + \frac{\sqrt{3}}{2})$
$= 5 g [\frac{2 + \sqrt{3}}{2}] = g (9.325)$
$F = 9.8 \times 9.325$
$F = 91.4 N$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

372171 A block of mass $m$ is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is $μ$ . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction

1 zero

2

2 μ mg

3

\frac{1}{2} μ mg

4

μ mg

Explanation:

A When lift freely fall, $N = 0$
We know that, $f = μ_{s} N$
$f = μ_{s} \times 0$
Force of friction $f = 0$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372172 The coefficient of friction between a hemispherical bowl and an insect is $\sqrt{0.44}$ and the radius of the bowl is $0.6 m$ . The maximum height to which an insect can crawl in the bowl will be

1

0.4 m

2

0.2 m

3

0.3 m

4

0.1 m

Explanation:

D $R = 0.6 cm$
$μ = \sqrt{0.44}$

From figure : $R = mg \cos θ$
$f = mg \sin θ$
$\Rightarrow \frac{f}{R} = \frac{mg \sin θ}{mg \cos θ} = \tan θ$
Also, $f = μ R \Rightarrow \frac{f}{R} = μ$
Hence, $\tan θ = μ$
$h = R - R \cos θ = R [1 - \cos θ]$
$= R [1 - \frac{1}{\sqrt{1 + μ^{2}}}]$
$\Rightarrow h = 0.6 [1 - \frac{1}{\sqrt{1 + 0.44}}] = 0.6 [\frac{1}{6}]$

$\tan θ = μ$
$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}$
$h = 0.1 cm$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372173 The rear side of truck is open and a box of mass $50 kg$ is placed at $3.2 m$ away from the open end. The coefficient of friction between the box. and the surface is 0.2 . If the truck starts from rest and moves on a straight road with acceleration $2.1 {ms}^{- 2}$ , then the box falls off the truck after a time $($ take $g = 10 {ms}^{- 2})$

1

4 s

2

\frac{8}{\sqrt{41}} s

3

8 s

4

\sqrt{8} s

Explanation:

C Acceleration due to pseudo force, $a_{p} = 2.1$ $m / \sec^{2}$
Net acceleration, $a = a_{p} - μ g$
Acceleration due to slipping between box and surface.
$a = 2.1 - (0.2 \times 10) = 0.1$
$t = \sqrt{\frac{2 s}{a}}$
Where, $a = 0.1$ and $s = 3.2$
$t = \sqrt{\frac{2 (3.2)}{0.1}} = \sqrt{64}$
$t = 8 \sec$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372174 A car is traveling along a curved road of radius $r$ . If the coefficient of friction between the tyres and the road is $μ$ , the car will skid if its speed exceeds

1

2 \sqrt{μ rg}

2

3 \sqrt{μ rg}

3

\sqrt{2 μ rg}

4

\sqrt{μ rg}

Explanation:

D
Let, Mass of car $= m$
Radius of curved road $= r$
For curved road-
Centripetal force $=$ Psudo force
$F_{c} = F_{r}$
$\frac{{mv}^{2}}{r} = μ mg$
Speed $v = \sqrt{μ r g}$

AP EMCET(Medical)-2010

LAWS OF MOTION (ADDITIONAL)

372175 A body of mass $10 kg$ is lying on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal and the coefficient of friction is 0.5 . The minimum force required to pull the body up the plane is

1

914 N

2

91.4 N

3

9.14 N

4

0.914 N

Explanation:

B
We know that,
$f_{k} = μ N$
$= μ mg \cos θ$
According to question-
$F = f_{k} + mg \sin θ$
$= μ mg \cos θ + mg \sin θ$
$= μ mg \cos 30^{\circ} + mg \sin 30^{\circ}$
$= 5 g \times \frac{\sqrt{3}}{2} + 10 g \times \frac{1}{2}$
$= 5 g + 5 g \times \frac{\sqrt{3}}{2}$
$= 5 g (1 + \frac{\sqrt{3}}{2})$
$= 5 g [\frac{2 + \sqrt{3}}{2}] = g (9.325)$
$F = 9.8 \times 9.325$
$F = 91.4 N$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

372171 A block of mass $m$ is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is $μ$ . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction

1 zero

2

2 μ mg

3

\frac{1}{2} μ mg

4

μ mg

Explanation:

A When lift freely fall, $N = 0$
We know that, $f = μ_{s} N$
$f = μ_{s} \times 0$
Force of friction $f = 0$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372172 The coefficient of friction between a hemispherical bowl and an insect is $\sqrt{0.44}$ and the radius of the bowl is $0.6 m$ . The maximum height to which an insect can crawl in the bowl will be

1

0.4 m

2

0.2 m

3

0.3 m

4

0.1 m

Explanation:

D $R = 0.6 cm$
$μ = \sqrt{0.44}$

From figure : $R = mg \cos θ$
$f = mg \sin θ$
$\Rightarrow \frac{f}{R} = \frac{mg \sin θ}{mg \cos θ} = \tan θ$
Also, $f = μ R \Rightarrow \frac{f}{R} = μ$
Hence, $\tan θ = μ$
$h = R - R \cos θ = R [1 - \cos θ]$
$= R [1 - \frac{1}{\sqrt{1 + μ^{2}}}]$
$\Rightarrow h = 0.6 [1 - \frac{1}{\sqrt{1 + 0.44}}] = 0.6 [\frac{1}{6}]$

$\tan θ = μ$
$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}$
$h = 0.1 cm$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372173 The rear side of truck is open and a box of mass $50 kg$ is placed at $3.2 m$ away from the open end. The coefficient of friction between the box. and the surface is 0.2 . If the truck starts from rest and moves on a straight road with acceleration $2.1 {ms}^{- 2}$ , then the box falls off the truck after a time $($ take $g = 10 {ms}^{- 2})$

1

4 s

2

\frac{8}{\sqrt{41}} s

3

8 s

4

\sqrt{8} s

Explanation:

C Acceleration due to pseudo force, $a_{p} = 2.1$ $m / \sec^{2}$
Net acceleration, $a = a_{p} - μ g$
Acceleration due to slipping between box and surface.
$a = 2.1 - (0.2 \times 10) = 0.1$
$t = \sqrt{\frac{2 s}{a}}$
Where, $a = 0.1$ and $s = 3.2$
$t = \sqrt{\frac{2 (3.2)}{0.1}} = \sqrt{64}$
$t = 8 \sec$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372174 A car is traveling along a curved road of radius $r$ . If the coefficient of friction between the tyres and the road is $μ$ , the car will skid if its speed exceeds

1

2 \sqrt{μ rg}

2

3 \sqrt{μ rg}

3

\sqrt{2 μ rg}

4

\sqrt{μ rg}

Explanation:

D
Let, Mass of car $= m$
Radius of curved road $= r$
For curved road-
Centripetal force $=$ Psudo force
$F_{c} = F_{r}$
$\frac{{mv}^{2}}{r} = μ mg$
Speed $v = \sqrt{μ r g}$

AP EMCET(Medical)-2010

LAWS OF MOTION (ADDITIONAL)

372175 A body of mass $10 kg$ is lying on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal and the coefficient of friction is 0.5 . The minimum force required to pull the body up the plane is

1

914 N

2

91.4 N

3

9.14 N

4

0.914 N

Explanation:

B
We know that,
$f_{k} = μ N$
$= μ mg \cos θ$
According to question-
$F = f_{k} + mg \sin θ$
$= μ mg \cos θ + mg \sin θ$
$= μ mg \cos 30^{\circ} + mg \sin 30^{\circ}$
$= 5 g \times \frac{\sqrt{3}}{2} + 10 g \times \frac{1}{2}$
$= 5 g + 5 g \times \frac{\sqrt{3}}{2}$
$= 5 g (1 + \frac{\sqrt{3}}{2})$
$= 5 g [\frac{2 + \sqrt{3}}{2}] = g (9.325)$
$F = 9.8 \times 9.325$
$F = 91.4 N$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

372171 A block of mass $m$ is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is $μ$ . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction

1 zero

2

2 μ mg

3

\frac{1}{2} μ mg

4

μ mg

Explanation:

A When lift freely fall, $N = 0$
We know that, $f = μ_{s} N$
$f = μ_{s} \times 0$
Force of friction $f = 0$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372172 The coefficient of friction between a hemispherical bowl and an insect is $\sqrt{0.44}$ and the radius of the bowl is $0.6 m$ . The maximum height to which an insect can crawl in the bowl will be

1

0.4 m

2

0.2 m

3

0.3 m

4

0.1 m

Explanation:

D $R = 0.6 cm$
$μ = \sqrt{0.44}$

From figure : $R = mg \cos θ$
$f = mg \sin θ$
$\Rightarrow \frac{f}{R} = \frac{mg \sin θ}{mg \cos θ} = \tan θ$
Also, $f = μ R \Rightarrow \frac{f}{R} = μ$
Hence, $\tan θ = μ$
$h = R - R \cos θ = R [1 - \cos θ]$
$= R [1 - \frac{1}{\sqrt{1 + μ^{2}}}]$
$\Rightarrow h = 0.6 [1 - \frac{1}{\sqrt{1 + 0.44}}] = 0.6 [\frac{1}{6}]$

$\tan θ = μ$
$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}$
$h = 0.1 cm$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372173 The rear side of truck is open and a box of mass $50 kg$ is placed at $3.2 m$ away from the open end. The coefficient of friction between the box. and the surface is 0.2 . If the truck starts from rest and moves on a straight road with acceleration $2.1 {ms}^{- 2}$ , then the box falls off the truck after a time $($ take $g = 10 {ms}^{- 2})$

1

4 s

2

\frac{8}{\sqrt{41}} s

3

8 s

4

\sqrt{8} s

Explanation:

C Acceleration due to pseudo force, $a_{p} = 2.1$ $m / \sec^{2}$
Net acceleration, $a = a_{p} - μ g$
Acceleration due to slipping between box and surface.
$a = 2.1 - (0.2 \times 10) = 0.1$
$t = \sqrt{\frac{2 s}{a}}$
Where, $a = 0.1$ and $s = 3.2$
$t = \sqrt{\frac{2 (3.2)}{0.1}} = \sqrt{64}$
$t = 8 \sec$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372174 A car is traveling along a curved road of radius $r$ . If the coefficient of friction between the tyres and the road is $μ$ , the car will skid if its speed exceeds

1

2 \sqrt{μ rg}

2

3 \sqrt{μ rg}

3

\sqrt{2 μ rg}

4

\sqrt{μ rg}

Explanation:

D
Let, Mass of car $= m$
Radius of curved road $= r$
For curved road-
Centripetal force $=$ Psudo force
$F_{c} = F_{r}$
$\frac{{mv}^{2}}{r} = μ mg$
Speed $v = \sqrt{μ r g}$

AP EMCET(Medical)-2010

LAWS OF MOTION (ADDITIONAL)

372175 A body of mass $10 kg$ is lying on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal and the coefficient of friction is 0.5 . The minimum force required to pull the body up the plane is

1

914 N

2

91.4 N

3

9.14 N

4

0.914 N

Explanation:

B
We know that,
$f_{k} = μ N$
$= μ mg \cos θ$
According to question-
$F = f_{k} + mg \sin θ$
$= μ mg \cos θ + mg \sin θ$
$= μ mg \cos 30^{\circ} + mg \sin 30^{\circ}$
$= 5 g \times \frac{\sqrt{3}}{2} + 10 g \times \frac{1}{2}$
$= 5 g + 5 g \times \frac{\sqrt{3}}{2}$
$= 5 g (1 + \frac{\sqrt{3}}{2})$
$= 5 g [\frac{2 + \sqrt{3}}{2}] = g (9.325)$
$F = 9.8 \times 9.325$
$F = 91.4 N$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

372171 A block of mass $m$ is placed on floor of a lift which is rough. The coefficient of friction between the block and the floor is $μ$ . When the lift falls freely, the block is pulled horizontally on the lift floor. The force of friction

1 zero

2

2 μ mg

3

\frac{1}{2} μ mg

4

μ mg

Explanation:

A When lift freely fall, $N = 0$
We know that, $f = μ_{s} N$
$f = μ_{s} \times 0$
Force of friction $f = 0$

AP EAMCET(Medical)-2012

LAWS OF MOTION (ADDITIONAL)

372172 The coefficient of friction between a hemispherical bowl and an insect is $\sqrt{0.44}$ and the radius of the bowl is $0.6 m$ . The maximum height to which an insect can crawl in the bowl will be

1

0.4 m

2

0.2 m

3

0.3 m

4

0.1 m

Explanation:

D $R = 0.6 cm$
$μ = \sqrt{0.44}$

From figure : $R = mg \cos θ$
$f = mg \sin θ$
$\Rightarrow \frac{f}{R} = \frac{mg \sin θ}{mg \cos θ} = \tan θ$
Also, $f = μ R \Rightarrow \frac{f}{R} = μ$
Hence, $\tan θ = μ$
$h = R - R \cos θ = R [1 - \cos θ]$
$= R [1 - \frac{1}{\sqrt{1 + μ^{2}}}]$
$\Rightarrow h = 0.6 [1 - \frac{1}{\sqrt{1 + 0.44}}] = 0.6 [\frac{1}{6}]$

$\tan θ = μ$
$\cos θ = \frac{1}{\sqrt{1 + μ^{2}}}$
$h = 0.1 cm$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372173 The rear side of truck is open and a box of mass $50 kg$ is placed at $3.2 m$ away from the open end. The coefficient of friction between the box. and the surface is 0.2 . If the truck starts from rest and moves on a straight road with acceleration $2.1 {ms}^{- 2}$ , then the box falls off the truck after a time $($ take $g = 10 {ms}^{- 2})$

1

4 s

2

\frac{8}{\sqrt{41}} s

3

8 s

4

\sqrt{8} s

Explanation:

C Acceleration due to pseudo force, $a_{p} = 2.1$ $m / \sec^{2}$
Net acceleration, $a = a_{p} - μ g$
Acceleration due to slipping between box and surface.
$a = 2.1 - (0.2 \times 10) = 0.1$
$t = \sqrt{\frac{2 s}{a}}$
Where, $a = 0.1$ and $s = 3.2$
$t = \sqrt{\frac{2 (3.2)}{0.1}} = \sqrt{64}$
$t = 8 \sec$

AP EMCET(Medical)-2011

LAWS OF MOTION (ADDITIONAL)

372174 A car is traveling along a curved road of radius $r$ . If the coefficient of friction between the tyres and the road is $μ$ , the car will skid if its speed exceeds

1

2 \sqrt{μ rg}

2

3 \sqrt{μ rg}

3

\sqrt{2 μ rg}

4

\sqrt{μ rg}

Explanation:

D
Let, Mass of car $= m$
Radius of curved road $= r$
For curved road-
Centripetal force $=$ Psudo force
$F_{c} = F_{r}$
$\frac{{mv}^{2}}{r} = μ mg$
Speed $v = \sqrt{μ r g}$

AP EMCET(Medical)-2010

LAWS OF MOTION (ADDITIONAL)

372175 A body of mass $10 kg$ is lying on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal and the coefficient of friction is 0.5 . The minimum force required to pull the body up the plane is

1

914 N

2

91.4 N

3

9.14 N

4

0.914 N

Explanation:

B
We know that,
$f_{k} = μ N$
$= μ mg \cos θ$
According to question-
$F = f_{k} + mg \sin θ$
$= μ mg \cos θ + mg \sin θ$
$= μ mg \cos 30^{\circ} + mg \sin 30^{\circ}$
$= 5 g \times \frac{\sqrt{3}}{2} + 10 g \times \frac{1}{2}$
$= 5 g + 5 g \times \frac{\sqrt{3}}{2}$
$= 5 g (1 + \frac{\sqrt{3}}{2})$
$= 5 g [\frac{2 + \sqrt{3}}{2}] = g (9.325)$
$F = 9.8 \times 9.325$
$F = 91.4 N$

AP EMCET(Medical)-2008

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