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PHXI05:LAWS OF MOTION

363390 A block placed on a horizontal surface is being pushed by a force $F$ making an angle $θ$ with the vertical. The coefficient of friction between block and surface is $μ$ . The force required to slide the block with uniform velocity on the floor is:

1

\frac{μ m g}{(\sin θ - μ \cos θ)}

2

\frac{(\sin θ - μ \cos θ)}{μ m g}

3

μ m g

4

N o n e o f t h e s e

Explanation:

For vertical equilibrium of the block:
$R = F \cos θ + m g (1)$
While for horizontal:
$F \sin θ = μ R (2)$
From (1) and (2).
$F \sin θ = μ (F \cos θ + m g)$
$\Rightarrow F = \frac{μ m g}{(\sin θ - μ \cos θ)}$
supporting img

PHXI05:LAWS OF MOTION

363391 As shown in the figure, a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ [Given $g = 10 m s^{- 2}$ ]
supporting img

1

35.7 N

2

33.3 N

3

20 N

4

25.2 N

Explanation:

Here,
$m = 10 k g, θ = 30^{\circ}, μ_{s} = 0.25, g = 10 m / s^{2}$
$\sum F_{y} = 0; F \sin 30^{\circ} + N = m g$
So, $N = m g - F \sin 30^{\circ}$
When the block is on the verge of motion,
$\sum F_{x} = 0; F \cos 30^{\circ} = f_{l} = μ N$
When block just start to move.
$F \cos 30^{\circ} = μ (m g - F \sin 30^{\circ})$
supporting img
$F (\cos 30^{\circ} + μ \sin 30^{\circ}) = μ m g$
$F [\frac{\sqrt{3}}{2} + 0.25 \times \frac{1}{2}] = 0.25 \times 10 \times 10$
$\Rightarrow F = 25.2 N$

JEE - 2023

PHXI05:LAWS OF MOTION

363392 Assertion :
The frictional force acts only when the bodies are in contact.
Reason :
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363393 A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ,$ the shortest distance in which the car can be stopped is

1

\frac{v_{0}^{2}}{2 μ g}

2

\frac{v_{0}}{μ g}

3

{(\frac{v_{0}}{μ g})}^{2}

4

\frac{v_{0}}{μ}

Explanation:

Retarding force $F = m a = μ m g$
$∴ a = μ g$
Now from equation of motion $v^{2} = v_{o}^{2} - 2 a s$
$\Rightarrow 0 = v_{o}^{2} - 2 a s \Rightarrow s = \frac{v_{o}^{2}}{2 a s} = \frac{v_{o}^{2}}{2 μ g}$
$∴ s = \frac{v_{0}^{2}}{2 μ g}$

PHXI05:LAWS OF MOTION

363394 To avoid slipping while walking on ice, one should take smaller steps because of the

1 Friction of ice is large

2 Larger normal reaction

3 Friction of ice is small

4 Smaller normal reaction

PHXI05:LAWS OF MOTION

363390 A block placed on a horizontal surface is being pushed by a force $F$ making an angle $θ$ with the vertical. The coefficient of friction between block and surface is $μ$ . The force required to slide the block with uniform velocity on the floor is:

1

\frac{μ m g}{(\sin θ - μ \cos θ)}

2

\frac{(\sin θ - μ \cos θ)}{μ m g}

3

μ m g

4

N o n e o f t h e s e

Explanation:

For vertical equilibrium of the block:
$R = F \cos θ + m g (1)$
While for horizontal:
$F \sin θ = μ R (2)$
From (1) and (2).
$F \sin θ = μ (F \cos θ + m g)$
$\Rightarrow F = \frac{μ m g}{(\sin θ - μ \cos θ)}$
supporting img

PHXI05:LAWS OF MOTION

363391 As shown in the figure, a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ [Given $g = 10 m s^{- 2}$ ]
supporting img

1

35.7 N

2

33.3 N

3

20 N

4

25.2 N

Explanation:

Here,
$m = 10 k g, θ = 30^{\circ}, μ_{s} = 0.25, g = 10 m / s^{2}$
$\sum F_{y} = 0; F \sin 30^{\circ} + N = m g$
So, $N = m g - F \sin 30^{\circ}$
When the block is on the verge of motion,
$\sum F_{x} = 0; F \cos 30^{\circ} = f_{l} = μ N$
When block just start to move.
$F \cos 30^{\circ} = μ (m g - F \sin 30^{\circ})$
supporting img
$F (\cos 30^{\circ} + μ \sin 30^{\circ}) = μ m g$
$F [\frac{\sqrt{3}}{2} + 0.25 \times \frac{1}{2}] = 0.25 \times 10 \times 10$
$\Rightarrow F = 25.2 N$

JEE - 2023

PHXI05:LAWS OF MOTION

363392 Assertion :
The frictional force acts only when the bodies are in contact.
Reason :
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363393 A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ,$ the shortest distance in which the car can be stopped is

1

\frac{v_{0}^{2}}{2 μ g}

2

\frac{v_{0}}{μ g}

3

{(\frac{v_{0}}{μ g})}^{2}

4

\frac{v_{0}}{μ}

Explanation:

Retarding force $F = m a = μ m g$
$∴ a = μ g$
Now from equation of motion $v^{2} = v_{o}^{2} - 2 a s$
$\Rightarrow 0 = v_{o}^{2} - 2 a s \Rightarrow s = \frac{v_{o}^{2}}{2 a s} = \frac{v_{o}^{2}}{2 μ g}$
$∴ s = \frac{v_{0}^{2}}{2 μ g}$

PHXI05:LAWS OF MOTION

363394 To avoid slipping while walking on ice, one should take smaller steps because of the

1 Friction of ice is large

2 Larger normal reaction

3 Friction of ice is small

4 Smaller normal reaction

PHXI05:LAWS OF MOTION

363390 A block placed on a horizontal surface is being pushed by a force $F$ making an angle $θ$ with the vertical. The coefficient of friction between block and surface is $μ$ . The force required to slide the block with uniform velocity on the floor is:

1

\frac{μ m g}{(\sin θ - μ \cos θ)}

2

\frac{(\sin θ - μ \cos θ)}{μ m g}

3

μ m g

4

N o n e o f t h e s e

Explanation:

For vertical equilibrium of the block:
$R = F \cos θ + m g (1)$
While for horizontal:
$F \sin θ = μ R (2)$
From (1) and (2).
$F \sin θ = μ (F \cos θ + m g)$
$\Rightarrow F = \frac{μ m g}{(\sin θ - μ \cos θ)}$
supporting img

PHXI05:LAWS OF MOTION

363391 As shown in the figure, a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ [Given $g = 10 m s^{- 2}$ ]
supporting img

1

35.7 N

2

33.3 N

3

20 N

4

25.2 N

Explanation:

Here,
$m = 10 k g, θ = 30^{\circ}, μ_{s} = 0.25, g = 10 m / s^{2}$
$\sum F_{y} = 0; F \sin 30^{\circ} + N = m g$
So, $N = m g - F \sin 30^{\circ}$
When the block is on the verge of motion,
$\sum F_{x} = 0; F \cos 30^{\circ} = f_{l} = μ N$
When block just start to move.
$F \cos 30^{\circ} = μ (m g - F \sin 30^{\circ})$
supporting img
$F (\cos 30^{\circ} + μ \sin 30^{\circ}) = μ m g$
$F [\frac{\sqrt{3}}{2} + 0.25 \times \frac{1}{2}] = 0.25 \times 10 \times 10$
$\Rightarrow F = 25.2 N$

JEE - 2023

PHXI05:LAWS OF MOTION

363392 Assertion :
The frictional force acts only when the bodies are in contact.
Reason :
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363393 A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ,$ the shortest distance in which the car can be stopped is

1

\frac{v_{0}^{2}}{2 μ g}

2

\frac{v_{0}}{μ g}

3

{(\frac{v_{0}}{μ g})}^{2}

4

\frac{v_{0}}{μ}

Explanation:

Retarding force $F = m a = μ m g$
$∴ a = μ g$
Now from equation of motion $v^{2} = v_{o}^{2} - 2 a s$
$\Rightarrow 0 = v_{o}^{2} - 2 a s \Rightarrow s = \frac{v_{o}^{2}}{2 a s} = \frac{v_{o}^{2}}{2 μ g}$
$∴ s = \frac{v_{0}^{2}}{2 μ g}$

PHXI05:LAWS OF MOTION

363394 To avoid slipping while walking on ice, one should take smaller steps because of the

1 Friction of ice is large

2 Larger normal reaction

3 Friction of ice is small

4 Smaller normal reaction

PHXI05:LAWS OF MOTION

363390 A block placed on a horizontal surface is being pushed by a force $F$ making an angle $θ$ with the vertical. The coefficient of friction between block and surface is $μ$ . The force required to slide the block with uniform velocity on the floor is:

1

\frac{μ m g}{(\sin θ - μ \cos θ)}

2

\frac{(\sin θ - μ \cos θ)}{μ m g}

3

μ m g

4

N o n e o f t h e s e

Explanation:

For vertical equilibrium of the block:
$R = F \cos θ + m g (1)$
While for horizontal:
$F \sin θ = μ R (2)$
From (1) and (2).
$F \sin θ = μ (F \cos θ + m g)$
$\Rightarrow F = \frac{μ m g}{(\sin θ - μ \cos θ)}$
supporting img

PHXI05:LAWS OF MOTION

363391 As shown in the figure, a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ [Given $g = 10 m s^{- 2}$ ]
supporting img

1

35.7 N

2

33.3 N

3

20 N

4

25.2 N

Explanation:

Here,
$m = 10 k g, θ = 30^{\circ}, μ_{s} = 0.25, g = 10 m / s^{2}$
$\sum F_{y} = 0; F \sin 30^{\circ} + N = m g$
So, $N = m g - F \sin 30^{\circ}$
When the block is on the verge of motion,
$\sum F_{x} = 0; F \cos 30^{\circ} = f_{l} = μ N$
When block just start to move.
$F \cos 30^{\circ} = μ (m g - F \sin 30^{\circ})$
supporting img
$F (\cos 30^{\circ} + μ \sin 30^{\circ}) = μ m g$
$F [\frac{\sqrt{3}}{2} + 0.25 \times \frac{1}{2}] = 0.25 \times 10 \times 10$
$\Rightarrow F = 25.2 N$

JEE - 2023

PHXI05:LAWS OF MOTION

363392 Assertion :
The frictional force acts only when the bodies are in contact.
Reason :
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363393 A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ,$ the shortest distance in which the car can be stopped is

1

\frac{v_{0}^{2}}{2 μ g}

2

\frac{v_{0}}{μ g}

3

{(\frac{v_{0}}{μ g})}^{2}

4

\frac{v_{0}}{μ}

Explanation:

Retarding force $F = m a = μ m g$
$∴ a = μ g$
Now from equation of motion $v^{2} = v_{o}^{2} - 2 a s$
$\Rightarrow 0 = v_{o}^{2} - 2 a s \Rightarrow s = \frac{v_{o}^{2}}{2 a s} = \frac{v_{o}^{2}}{2 μ g}$
$∴ s = \frac{v_{0}^{2}}{2 μ g}$

PHXI05:LAWS OF MOTION

363394 To avoid slipping while walking on ice, one should take smaller steps because of the

1 Friction of ice is large

2 Larger normal reaction

3 Friction of ice is small

4 Smaller normal reaction

PHXI05:LAWS OF MOTION

363390 A block placed on a horizontal surface is being pushed by a force $F$ making an angle $θ$ with the vertical. The coefficient of friction between block and surface is $μ$ . The force required to slide the block with uniform velocity on the floor is:

1

\frac{μ m g}{(\sin θ - μ \cos θ)}

2

\frac{(\sin θ - μ \cos θ)}{μ m g}

3

μ m g

4

N o n e o f t h e s e

Explanation:

For vertical equilibrium of the block:
$R = F \cos θ + m g (1)$
While for horizontal:
$F \sin θ = μ R (2)$
From (1) and (2).
$F \sin θ = μ (F \cos θ + m g)$
$\Rightarrow F = \frac{μ m g}{(\sin θ - μ \cos θ)}$
supporting img

PHXI05:LAWS OF MOTION

363391 As shown in the figure, a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ [Given $g = 10 m s^{- 2}$ ]
supporting img

1

35.7 N

2

33.3 N

3

20 N

4

25.2 N

Explanation:

Here,
$m = 10 k g, θ = 30^{\circ}, μ_{s} = 0.25, g = 10 m / s^{2}$
$\sum F_{y} = 0; F \sin 30^{\circ} + N = m g$
So, $N = m g - F \sin 30^{\circ}$
When the block is on the verge of motion,
$\sum F_{x} = 0; F \cos 30^{\circ} = f_{l} = μ N$
When block just start to move.
$F \cos 30^{\circ} = μ (m g - F \sin 30^{\circ})$
supporting img
$F (\cos 30^{\circ} + μ \sin 30^{\circ}) = μ m g$
$F [\frac{\sqrt{3}}{2} + 0.25 \times \frac{1}{2}] = 0.25 \times 10 \times 10$
$\Rightarrow F = 25.2 N$

JEE - 2023

PHXI05:LAWS OF MOTION

363392 Assertion :
The frictional force acts only when the bodies are in contact.
Reason :
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363393 A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ,$ the shortest distance in which the car can be stopped is

1

\frac{v_{0}^{2}}{2 μ g}

2

\frac{v_{0}}{μ g}

3

{(\frac{v_{0}}{μ g})}^{2}

4

\frac{v_{0}}{μ}

Explanation:

Retarding force $F = m a = μ m g$
$∴ a = μ g$
Now from equation of motion $v^{2} = v_{o}^{2} - 2 a s$
$\Rightarrow 0 = v_{o}^{2} - 2 a s \Rightarrow s = \frac{v_{o}^{2}}{2 a s} = \frac{v_{o}^{2}}{2 μ g}$
$∴ s = \frac{v_{0}^{2}}{2 μ g}$

PHXI05:LAWS OF MOTION

363394 To avoid slipping while walking on ice, one should take smaller steps because of the

1 Friction of ice is large

2 Larger normal reaction

3 Friction of ice is small

4 Smaller normal reaction

Explanation:

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