QBManager

LAWS OF MOTION (ADDITIONAL)

372090 When the two surfaces are coated with the lubricant, then they will:

1 slide upon each other

2 stick to each other

3 roll upon each other

4 none of these

AIIMS-2001

LAWS OF MOTION (ADDITIONAL)

372091 If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at $72 km / h$ ?

1

50 m

2

60 m

3

40.8 m

4

80.16 m

Explanation:

C Given,
$v = 0$
$u = 72 km / h = 20 m / s$
$g = 9.8 m / s^{2}$
Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction $(f_{s})$ working between tyres and road, so frictional force cause the retardation in velocity of automobile.

Hence,
$F = f_{s} = μ R$
$= μ . mg$
Where $m$ is the mass of automobile
We know that,
$F = ma$
$μ mg = ma$
$a = μ g$
$a = 0.5 g$
Retardation is $0.5 g$
Let automobile stops at a distance $x$ , then from third equation of motion-
$v^{2} = u^{2} - 2 as$
$0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$
$x = \frac{20 \times 20}{2 \times 0.5 \times 9.8} = 40.8 m$
$∴ 0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$

BCECE-2007

LAWS OF MOTION (ADDITIONAL)

372092 In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass $m$ is $f$ , then its acceleration in terms of the force $F$ will be eaual to :

1

(F - f) / m

2

(\frac{F}{2} - f) / m

3

F / m

4 none of these

Explanation:

B As the pulley is mass-less and frictionless, net force on pulley will be zero.

According to question -

$\frac{F}{2} - f = ma$
$a = \frac{(\frac{F}{2} - f)}{m}$

BCECE-2006

LAWS OF MOTION (ADDITIONAL)

372093 A body is moving along a rough horizontal surface with an initial velocity $6 m / s$ . If the body comes to rest after travelling a distance 9 $m$ , then the coefficient of sliding friction will be:

1 0.4

2 0.2

3 0.6

4 0.8

Explanation:

B Given,
Initial velocity $(u) = 6 m / s, s = 9 m, μ_{s} =$ ?
We know that,
$F = μ_{s} \cdot N$
$= μ_{s} \cdot mg$
If acceleration of body is a
Then,
$F = μ_{s} \cdot mg$
$m \cdot a = μ_{s} \cdot mg$
$μ_{s} = \frac{a}{g}$
Where, $μ_{s}$ is sliding friction.
Apply equation of motion -
$v^{2} = u^{2} + 2 a s$
$0 = (6)^{2} + 2 a \times 9$
$- 36 = 18 a$
$a = - 2 m / s^{2}$ (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is -
$μ_{s} = \frac{a}{g} = \frac{2}{10} = 0.2$

BCECE-2004

LAWS OF MOTION (ADDITIONAL)

372090 When the two surfaces are coated with the lubricant, then they will:

1 slide upon each other

2 stick to each other

3 roll upon each other

4 none of these

AIIMS-2001

LAWS OF MOTION (ADDITIONAL)

372091 If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at $72 km / h$ ?

1

50 m

2

60 m

3

40.8 m

4

80.16 m

Explanation:

C Given,
$v = 0$
$u = 72 km / h = 20 m / s$
$g = 9.8 m / s^{2}$
Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction $(f_{s})$ working between tyres and road, so frictional force cause the retardation in velocity of automobile.

Hence,
$F = f_{s} = μ R$
$= μ . mg$
Where $m$ is the mass of automobile
We know that,
$F = ma$
$μ mg = ma$
$a = μ g$
$a = 0.5 g$
Retardation is $0.5 g$
Let automobile stops at a distance $x$ , then from third equation of motion-
$v^{2} = u^{2} - 2 as$
$0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$
$x = \frac{20 \times 20}{2 \times 0.5 \times 9.8} = 40.8 m$
$∴ 0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$

BCECE-2007

LAWS OF MOTION (ADDITIONAL)

372092 In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass $m$ is $f$ , then its acceleration in terms of the force $F$ will be eaual to :

1

(F - f) / m

2

(\frac{F}{2} - f) / m

3

F / m

4 none of these

Explanation:

B As the pulley is mass-less and frictionless, net force on pulley will be zero.

According to question -

$\frac{F}{2} - f = ma$
$a = \frac{(\frac{F}{2} - f)}{m}$

BCECE-2006

LAWS OF MOTION (ADDITIONAL)

372093 A body is moving along a rough horizontal surface with an initial velocity $6 m / s$ . If the body comes to rest after travelling a distance 9 $m$ , then the coefficient of sliding friction will be:

1 0.4

2 0.2

3 0.6

4 0.8

Explanation:

B Given,
Initial velocity $(u) = 6 m / s, s = 9 m, μ_{s} =$ ?
We know that,
$F = μ_{s} \cdot N$
$= μ_{s} \cdot mg$
If acceleration of body is a
Then,
$F = μ_{s} \cdot mg$
$m \cdot a = μ_{s} \cdot mg$
$μ_{s} = \frac{a}{g}$
Where, $μ_{s}$ is sliding friction.
Apply equation of motion -
$v^{2} = u^{2} + 2 a s$
$0 = (6)^{2} + 2 a \times 9$
$- 36 = 18 a$
$a = - 2 m / s^{2}$ (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is -
$μ_{s} = \frac{a}{g} = \frac{2}{10} = 0.2$

BCECE-2004

LAWS OF MOTION (ADDITIONAL)

372090 When the two surfaces are coated with the lubricant, then they will:

1 slide upon each other

2 stick to each other

3 roll upon each other

4 none of these

AIIMS-2001

LAWS OF MOTION (ADDITIONAL)

372091 If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at $72 km / h$ ?

1

50 m

2

60 m

3

40.8 m

4

80.16 m

Explanation:

C Given,
$v = 0$
$u = 72 km / h = 20 m / s$
$g = 9.8 m / s^{2}$
Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction $(f_{s})$ working between tyres and road, so frictional force cause the retardation in velocity of automobile.

Hence,
$F = f_{s} = μ R$
$= μ . mg$
Where $m$ is the mass of automobile
We know that,
$F = ma$
$μ mg = ma$
$a = μ g$
$a = 0.5 g$
Retardation is $0.5 g$
Let automobile stops at a distance $x$ , then from third equation of motion-
$v^{2} = u^{2} - 2 as$
$0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$
$x = \frac{20 \times 20}{2 \times 0.5 \times 9.8} = 40.8 m$
$∴ 0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$

BCECE-2007

LAWS OF MOTION (ADDITIONAL)

372092 In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass $m$ is $f$ , then its acceleration in terms of the force $F$ will be eaual to :

1

(F - f) / m

2

(\frac{F}{2} - f) / m

3

F / m

4 none of these

Explanation:

B As the pulley is mass-less and frictionless, net force on pulley will be zero.

According to question -

$\frac{F}{2} - f = ma$
$a = \frac{(\frac{F}{2} - f)}{m}$

BCECE-2006

LAWS OF MOTION (ADDITIONAL)

372093 A body is moving along a rough horizontal surface with an initial velocity $6 m / s$ . If the body comes to rest after travelling a distance 9 $m$ , then the coefficient of sliding friction will be:

1 0.4

2 0.2

3 0.6

4 0.8

Explanation:

B Given,
Initial velocity $(u) = 6 m / s, s = 9 m, μ_{s} =$ ?
We know that,
$F = μ_{s} \cdot N$
$= μ_{s} \cdot mg$
If acceleration of body is a
Then,
$F = μ_{s} \cdot mg$
$m \cdot a = μ_{s} \cdot mg$
$μ_{s} = \frac{a}{g}$
Where, $μ_{s}$ is sliding friction.
Apply equation of motion -
$v^{2} = u^{2} + 2 a s$
$0 = (6)^{2} + 2 a \times 9$
$- 36 = 18 a$
$a = - 2 m / s^{2}$ (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is -
$μ_{s} = \frac{a}{g} = \frac{2}{10} = 0.2$

BCECE-2004

LAWS OF MOTION (ADDITIONAL)

372090 When the two surfaces are coated with the lubricant, then they will:

1 slide upon each other

2 stick to each other

3 roll upon each other

4 none of these

AIIMS-2001

LAWS OF MOTION (ADDITIONAL)

372091 If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at $72 km / h$ ?

1

50 m

2

60 m

3

40.8 m

4

80.16 m

Explanation:

C Given,
$v = 0$
$u = 72 km / h = 20 m / s$
$g = 9.8 m / s^{2}$
Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction $(f_{s})$ working between tyres and road, so frictional force cause the retardation in velocity of automobile.

Hence,
$F = f_{s} = μ R$
$= μ . mg$
Where $m$ is the mass of automobile
We know that,
$F = ma$
$μ mg = ma$
$a = μ g$
$a = 0.5 g$
Retardation is $0.5 g$
Let automobile stops at a distance $x$ , then from third equation of motion-
$v^{2} = u^{2} - 2 as$
$0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$
$x = \frac{20 \times 20}{2 \times 0.5 \times 9.8} = 40.8 m$
$∴ 0^{2} = (20)^{2} - 2 \times (0.5 \times 9.8) x$

BCECE-2007

LAWS OF MOTION (ADDITIONAL)

372092 In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass $m$ is $f$ , then its acceleration in terms of the force $F$ will be eaual to :

1

(F - f) / m

2

(\frac{F}{2} - f) / m

3

F / m

4 none of these

Explanation:

B As the pulley is mass-less and frictionless, net force on pulley will be zero.

According to question -

$\frac{F}{2} - f = ma$
$a = \frac{(\frac{F}{2} - f)}{m}$

BCECE-2006

LAWS OF MOTION (ADDITIONAL)

372093 A body is moving along a rough horizontal surface with an initial velocity $6 m / s$ . If the body comes to rest after travelling a distance 9 $m$ , then the coefficient of sliding friction will be:

1 0.4

2 0.2

3 0.6

4 0.8

Explanation:

B Given,
Initial velocity $(u) = 6 m / s, s = 9 m, μ_{s} =$ ?
We know that,
$F = μ_{s} \cdot N$
$= μ_{s} \cdot mg$
If acceleration of body is a
Then,
$F = μ_{s} \cdot mg$
$m \cdot a = μ_{s} \cdot mg$
$μ_{s} = \frac{a}{g}$
Where, $μ_{s}$ is sliding friction.
Apply equation of motion -
$v^{2} = u^{2} + 2 a s$
$0 = (6)^{2} + 2 a \times 9$
$- 36 = 18 a$
$a = - 2 m / s^{2}$ (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is -
$μ_{s} = \frac{a}{g} = \frac{2}{10} = 0.2$

BCECE-2004

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