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Laws of Motion

146162 A body starts sliding down from the top of an inclined plane at an angle \(\theta\) with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\mu\).
If the body comes to rest at the bottom of the plane then the value of \(\mu\) is

1 \(\frac{\tan \theta}{2}\)

2 \(\frac{3 \tan \theta}{2}\)

3 \(\tan \theta\)

4 \(2 \tan \theta\)

Explanation:

D
Let mass of body \(=\mathrm{m}\)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{x}\)
Plane coefficient of friction,
\(\mu_{\mathrm{AB}}=0, \mu_{\mathrm{BC}}=\frac{\mu}{2}, \mu_{\mathrm{CD}}=\mu\)
For inclined plane \(A B\),
Friction force \(\left(\mathrm{f}_{\mathrm{AB}}\right)=\mu \mathrm{N}=0 \times \mathrm{N}=0\)
\(\mathrm{F}_{\text {net }}\) equation,
\(\mathrm{mg} \sin \theta-0=\mathrm{ma}_{\mathrm{AB}}\)
\(\mathrm{a}_{\mathrm{AB}}=\mathrm{g} \sin \theta\)
Applying third equation of motion,
\(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\)
\(\mathrm{v}_{\mathrm{AB}}^{2}=2 \times \mathrm{g} \sin \theta \times \mathrm{x}=2 \mathrm{gx} \sin \theta\)
For inclined plane \(\mathrm{BC}\),
Friction force \(\left(\mathrm{f}_{\mathrm{BC}}\right)=(\mu / 2) \mathrm{mg} \cos \theta\)
So, net force equation-
\(\mathrm{mg} \sin \theta-(\mu / 2) \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{BC}}\)
\(\mathrm{a}_{\mathrm{BC}}=\mathrm{g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{BC}}^{2}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{BC}} \mathrm{x} \quad\left[\because \mathrm{u}_{\mathrm{BC}}=\mathrm{v}_{\mathrm{AB}}\right]\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=2 \mathrm{gx} \sin \theta+2\left(\mathrm{~g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\right) \mathrm{x}\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=4 \mathrm{gx} \sin \theta-\mu \mathrm{gx} \cos \theta\)
For CD,
Friction force \(\left(\mathrm{f}_{\mathrm{CD}}\right)=\mu \mathrm{mg} \cos \theta\)
So, net force equation,
\(\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{CD}}\)
\(\mathrm{a}_{\mathrm{CD}}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{CD}}^{2}=\mathrm{u}_{\mathrm{CD}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x} \quad\left(\because \mathrm{u}_{\mathrm{CD}}=\mathrm{v}_{\mathrm{BC}}\right)\)
\(0=6 g x \sin \theta-3 \mu g x \cos \theta\)
\(6 g x \sin \theta=3 \mu g x \cos \theta\)
\(\mu=\frac{6 g x \sin \theta}{3 g x \cos \theta}\)
\(\mu=2 \tan \theta\)

AP EAMCET (22.04.2019) Shift-II

Laws of Motion

146163 A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontal by with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{2}{9}\)

2 \(\frac{5}{12}\)

3 \(\frac{1}{5}\)

4 \(\frac{2}{5}\)

AP EAMCET (21.04.2019) Shift-I

Laws of Motion

146164 The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is \(60^{\circ}\), then the coefficient of friction is

1 \(\frac{1}{3}\)

2 \(\frac{1}{\sqrt{2}}\)

3 \(\frac{1}{\sqrt{3}}\)

4 \(\frac{1}{2}\)

AP EAMCET (20.04.2019) Shift-II

Laws of Motion

146165 \(\quad\) The maximum value of the applied force \(F\) such that the block as shown in the arrangement does not move is
(Acceleration due to gravity, \(g=10 \mathbf{m s}^{-2}\) )

1 \(20 \mathrm{~N}\)

2 \(15 \mathrm{~N}\)

3 \(25 \mathrm{~N}\)

4 \(10 \mathrm{~N}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146166 A body of mass \(2 \mathrm{~kg}\) is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 \(\mathrm{N}\), then the frictional force acting on the body will be \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)

1 \(8 \mathrm{~N}\)

2 \(10 \mathrm{~N}\)

3 \(20 \mathrm{~N}\)

4 \(2.5 \mathrm{~N}\)

AIIMS-26.05.2018(M)

Laws of Motion

146162 A body starts sliding down from the top of an inclined plane at an angle \(\theta\) with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\mu\).
If the body comes to rest at the bottom of the plane then the value of \(\mu\) is

1 \(\frac{\tan \theta}{2}\)

2 \(\frac{3 \tan \theta}{2}\)

3 \(\tan \theta\)

4 \(2 \tan \theta\)

Explanation:

D
Let mass of body \(=\mathrm{m}\)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{x}\)
Plane coefficient of friction,
\(\mu_{\mathrm{AB}}=0, \mu_{\mathrm{BC}}=\frac{\mu}{2}, \mu_{\mathrm{CD}}=\mu\)
For inclined plane \(A B\),
Friction force \(\left(\mathrm{f}_{\mathrm{AB}}\right)=\mu \mathrm{N}=0 \times \mathrm{N}=0\)
\(\mathrm{F}_{\text {net }}\) equation,
\(\mathrm{mg} \sin \theta-0=\mathrm{ma}_{\mathrm{AB}}\)
\(\mathrm{a}_{\mathrm{AB}}=\mathrm{g} \sin \theta\)
Applying third equation of motion,
\(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\)
\(\mathrm{v}_{\mathrm{AB}}^{2}=2 \times \mathrm{g} \sin \theta \times \mathrm{x}=2 \mathrm{gx} \sin \theta\)
For inclined plane \(\mathrm{BC}\),
Friction force \(\left(\mathrm{f}_{\mathrm{BC}}\right)=(\mu / 2) \mathrm{mg} \cos \theta\)
So, net force equation-
\(\mathrm{mg} \sin \theta-(\mu / 2) \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{BC}}\)
\(\mathrm{a}_{\mathrm{BC}}=\mathrm{g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{BC}}^{2}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{BC}} \mathrm{x} \quad\left[\because \mathrm{u}_{\mathrm{BC}}=\mathrm{v}_{\mathrm{AB}}\right]\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=2 \mathrm{gx} \sin \theta+2\left(\mathrm{~g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\right) \mathrm{x}\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=4 \mathrm{gx} \sin \theta-\mu \mathrm{gx} \cos \theta\)
For CD,
Friction force \(\left(\mathrm{f}_{\mathrm{CD}}\right)=\mu \mathrm{mg} \cos \theta\)
So, net force equation,
\(\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{CD}}\)
\(\mathrm{a}_{\mathrm{CD}}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{CD}}^{2}=\mathrm{u}_{\mathrm{CD}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x} \quad\left(\because \mathrm{u}_{\mathrm{CD}}=\mathrm{v}_{\mathrm{BC}}\right)\)
\(0=6 g x \sin \theta-3 \mu g x \cos \theta\)
\(6 g x \sin \theta=3 \mu g x \cos \theta\)
\(\mu=\frac{6 g x \sin \theta}{3 g x \cos \theta}\)
\(\mu=2 \tan \theta\)

AP EAMCET (22.04.2019) Shift-II

Laws of Motion

146163 A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontal by with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{2}{9}\)

2 \(\frac{5}{12}\)

3 \(\frac{1}{5}\)

4 \(\frac{2}{5}\)

AP EAMCET (21.04.2019) Shift-I

Laws of Motion

146164 The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is \(60^{\circ}\), then the coefficient of friction is

1 \(\frac{1}{3}\)

2 \(\frac{1}{\sqrt{2}}\)

3 \(\frac{1}{\sqrt{3}}\)

4 \(\frac{1}{2}\)

AP EAMCET (20.04.2019) Shift-II

Laws of Motion

146165 \(\quad\) The maximum value of the applied force \(F\) such that the block as shown in the arrangement does not move is
(Acceleration due to gravity, \(g=10 \mathbf{m s}^{-2}\) )

1 \(20 \mathrm{~N}\)

2 \(15 \mathrm{~N}\)

3 \(25 \mathrm{~N}\)

4 \(10 \mathrm{~N}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146166 A body of mass \(2 \mathrm{~kg}\) is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 \(\mathrm{N}\), then the frictional force acting on the body will be \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)

1 \(8 \mathrm{~N}\)

2 \(10 \mathrm{~N}\)

3 \(20 \mathrm{~N}\)

4 \(2.5 \mathrm{~N}\)

AIIMS-26.05.2018(M)

Laws of Motion

146162 A body starts sliding down from the top of an inclined plane at an angle \(\theta\) with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\mu\).
If the body comes to rest at the bottom of the plane then the value of \(\mu\) is

1 \(\frac{\tan \theta}{2}\)

2 \(\frac{3 \tan \theta}{2}\)

3 \(\tan \theta\)

4 \(2 \tan \theta\)

Explanation:

D
Let mass of body \(=\mathrm{m}\)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{x}\)
Plane coefficient of friction,
\(\mu_{\mathrm{AB}}=0, \mu_{\mathrm{BC}}=\frac{\mu}{2}, \mu_{\mathrm{CD}}=\mu\)
For inclined plane \(A B\),
Friction force \(\left(\mathrm{f}_{\mathrm{AB}}\right)=\mu \mathrm{N}=0 \times \mathrm{N}=0\)
\(\mathrm{F}_{\text {net }}\) equation,
\(\mathrm{mg} \sin \theta-0=\mathrm{ma}_{\mathrm{AB}}\)
\(\mathrm{a}_{\mathrm{AB}}=\mathrm{g} \sin \theta\)
Applying third equation of motion,
\(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\)
\(\mathrm{v}_{\mathrm{AB}}^{2}=2 \times \mathrm{g} \sin \theta \times \mathrm{x}=2 \mathrm{gx} \sin \theta\)
For inclined plane \(\mathrm{BC}\),
Friction force \(\left(\mathrm{f}_{\mathrm{BC}}\right)=(\mu / 2) \mathrm{mg} \cos \theta\)
So, net force equation-
\(\mathrm{mg} \sin \theta-(\mu / 2) \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{BC}}\)
\(\mathrm{a}_{\mathrm{BC}}=\mathrm{g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{BC}}^{2}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{BC}} \mathrm{x} \quad\left[\because \mathrm{u}_{\mathrm{BC}}=\mathrm{v}_{\mathrm{AB}}\right]\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=2 \mathrm{gx} \sin \theta+2\left(\mathrm{~g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\right) \mathrm{x}\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=4 \mathrm{gx} \sin \theta-\mu \mathrm{gx} \cos \theta\)
For CD,
Friction force \(\left(\mathrm{f}_{\mathrm{CD}}\right)=\mu \mathrm{mg} \cos \theta\)
So, net force equation,
\(\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{CD}}\)
\(\mathrm{a}_{\mathrm{CD}}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{CD}}^{2}=\mathrm{u}_{\mathrm{CD}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x} \quad\left(\because \mathrm{u}_{\mathrm{CD}}=\mathrm{v}_{\mathrm{BC}}\right)\)
\(0=6 g x \sin \theta-3 \mu g x \cos \theta\)
\(6 g x \sin \theta=3 \mu g x \cos \theta\)
\(\mu=\frac{6 g x \sin \theta}{3 g x \cos \theta}\)
\(\mu=2 \tan \theta\)

AP EAMCET (22.04.2019) Shift-II

Laws of Motion

146163 A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontal by with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{2}{9}\)

2 \(\frac{5}{12}\)

3 \(\frac{1}{5}\)

4 \(\frac{2}{5}\)

AP EAMCET (21.04.2019) Shift-I

Laws of Motion

146164 The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is \(60^{\circ}\), then the coefficient of friction is

1 \(\frac{1}{3}\)

2 \(\frac{1}{\sqrt{2}}\)

3 \(\frac{1}{\sqrt{3}}\)

4 \(\frac{1}{2}\)

AP EAMCET (20.04.2019) Shift-II

Laws of Motion

146165 \(\quad\) The maximum value of the applied force \(F\) such that the block as shown in the arrangement does not move is
(Acceleration due to gravity, \(g=10 \mathbf{m s}^{-2}\) )

1 \(20 \mathrm{~N}\)

2 \(15 \mathrm{~N}\)

3 \(25 \mathrm{~N}\)

4 \(10 \mathrm{~N}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146166 A body of mass \(2 \mathrm{~kg}\) is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 \(\mathrm{N}\), then the frictional force acting on the body will be \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)

1 \(8 \mathrm{~N}\)

2 \(10 \mathrm{~N}\)

3 \(20 \mathrm{~N}\)

4 \(2.5 \mathrm{~N}\)

AIIMS-26.05.2018(M)

Laws of Motion

146162 A body starts sliding down from the top of an inclined plane at an angle \(\theta\) with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\mu\).
If the body comes to rest at the bottom of the plane then the value of \(\mu\) is

1 \(\frac{\tan \theta}{2}\)

2 \(\frac{3 \tan \theta}{2}\)

3 \(\tan \theta\)

4 \(2 \tan \theta\)

Explanation:

D
Let mass of body \(=\mathrm{m}\)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{x}\)
Plane coefficient of friction,
\(\mu_{\mathrm{AB}}=0, \mu_{\mathrm{BC}}=\frac{\mu}{2}, \mu_{\mathrm{CD}}=\mu\)
For inclined plane \(A B\),
Friction force \(\left(\mathrm{f}_{\mathrm{AB}}\right)=\mu \mathrm{N}=0 \times \mathrm{N}=0\)
\(\mathrm{F}_{\text {net }}\) equation,
\(\mathrm{mg} \sin \theta-0=\mathrm{ma}_{\mathrm{AB}}\)
\(\mathrm{a}_{\mathrm{AB}}=\mathrm{g} \sin \theta\)
Applying third equation of motion,
\(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\)
\(\mathrm{v}_{\mathrm{AB}}^{2}=2 \times \mathrm{g} \sin \theta \times \mathrm{x}=2 \mathrm{gx} \sin \theta\)
For inclined plane \(\mathrm{BC}\),
Friction force \(\left(\mathrm{f}_{\mathrm{BC}}\right)=(\mu / 2) \mathrm{mg} \cos \theta\)
So, net force equation-
\(\mathrm{mg} \sin \theta-(\mu / 2) \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{BC}}\)
\(\mathrm{a}_{\mathrm{BC}}=\mathrm{g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{BC}}^{2}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{BC}} \mathrm{x} \quad\left[\because \mathrm{u}_{\mathrm{BC}}=\mathrm{v}_{\mathrm{AB}}\right]\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=2 \mathrm{gx} \sin \theta+2\left(\mathrm{~g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\right) \mathrm{x}\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=4 \mathrm{gx} \sin \theta-\mu \mathrm{gx} \cos \theta\)
For CD,
Friction force \(\left(\mathrm{f}_{\mathrm{CD}}\right)=\mu \mathrm{mg} \cos \theta\)
So, net force equation,
\(\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{CD}}\)
\(\mathrm{a}_{\mathrm{CD}}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{CD}}^{2}=\mathrm{u}_{\mathrm{CD}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x} \quad\left(\because \mathrm{u}_{\mathrm{CD}}=\mathrm{v}_{\mathrm{BC}}\right)\)
\(0=6 g x \sin \theta-3 \mu g x \cos \theta\)
\(6 g x \sin \theta=3 \mu g x \cos \theta\)
\(\mu=\frac{6 g x \sin \theta}{3 g x \cos \theta}\)
\(\mu=2 \tan \theta\)

AP EAMCET (22.04.2019) Shift-II

Laws of Motion

146163 A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontal by with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{2}{9}\)

2 \(\frac{5}{12}\)

3 \(\frac{1}{5}\)

4 \(\frac{2}{5}\)

AP EAMCET (21.04.2019) Shift-I

Laws of Motion

146164 The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is \(60^{\circ}\), then the coefficient of friction is

1 \(\frac{1}{3}\)

2 \(\frac{1}{\sqrt{2}}\)

3 \(\frac{1}{\sqrt{3}}\)

4 \(\frac{1}{2}\)

AP EAMCET (20.04.2019) Shift-II

Laws of Motion

146165 \(\quad\) The maximum value of the applied force \(F\) such that the block as shown in the arrangement does not move is
(Acceleration due to gravity, \(g=10 \mathbf{m s}^{-2}\) )

1 \(20 \mathrm{~N}\)

2 \(15 \mathrm{~N}\)

3 \(25 \mathrm{~N}\)

4 \(10 \mathrm{~N}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146166 A body of mass \(2 \mathrm{~kg}\) is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 \(\mathrm{N}\), then the frictional force acting on the body will be \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)

1 \(8 \mathrm{~N}\)

2 \(10 \mathrm{~N}\)

3 \(20 \mathrm{~N}\)

4 \(2.5 \mathrm{~N}\)

AIIMS-26.05.2018(M)

Laws of Motion

146162 A body starts sliding down from the top of an inclined plane at an angle \(\theta\) with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\frac{\mu}{2}\) and the last one third has coefficient of friction \(\mu\).
If the body comes to rest at the bottom of the plane then the value of \(\mu\) is

1 \(\frac{\tan \theta}{2}\)

2 \(\frac{3 \tan \theta}{2}\)

3 \(\tan \theta\)

4 \(2 \tan \theta\)

Explanation:

D
Let mass of body \(=\mathrm{m}\)
\(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{x}\)
Plane coefficient of friction,
\(\mu_{\mathrm{AB}}=0, \mu_{\mathrm{BC}}=\frac{\mu}{2}, \mu_{\mathrm{CD}}=\mu\)
For inclined plane \(A B\),
Friction force \(\left(\mathrm{f}_{\mathrm{AB}}\right)=\mu \mathrm{N}=0 \times \mathrm{N}=0\)
\(\mathrm{F}_{\text {net }}\) equation,
\(\mathrm{mg} \sin \theta-0=\mathrm{ma}_{\mathrm{AB}}\)
\(\mathrm{a}_{\mathrm{AB}}=\mathrm{g} \sin \theta\)
Applying third equation of motion,
\(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\)
\(\mathrm{v}_{\mathrm{AB}}^{2}=2 \times \mathrm{g} \sin \theta \times \mathrm{x}=2 \mathrm{gx} \sin \theta\)
For inclined plane \(\mathrm{BC}\),
Friction force \(\left(\mathrm{f}_{\mathrm{BC}}\right)=(\mu / 2) \mathrm{mg} \cos \theta\)
So, net force equation-
\(\mathrm{mg} \sin \theta-(\mu / 2) \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{BC}}\)
\(\mathrm{a}_{\mathrm{BC}}=\mathrm{g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{BC}}^{2}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{BC}} \mathrm{x} \quad\left[\because \mathrm{u}_{\mathrm{BC}}=\mathrm{v}_{\mathrm{AB}}\right]\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=2 \mathrm{gx} \sin \theta+2\left(\mathrm{~g} \sin \theta-\frac{\mu}{2} \mathrm{~g} \cos \theta\right) \mathrm{x}\)
\(\mathrm{v}_{\mathrm{BC}}^{2}=4 \mathrm{gx} \sin \theta-\mu \mathrm{gx} \cos \theta\)
For CD,
Friction force \(\left(\mathrm{f}_{\mathrm{CD}}\right)=\mu \mathrm{mg} \cos \theta\)
So, net force equation,
\(\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=\mathrm{ma}_{\mathrm{CD}}\)
\(\mathrm{a}_{\mathrm{CD}}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta\)
Applying third equation of motion,
\(\mathrm{v}_{\mathrm{CD}}^{2}=\mathrm{u}_{\mathrm{CD}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x}=\mathrm{u}_{\mathrm{BC}}^{2}+2 \mathrm{a}_{\mathrm{CD}} \mathrm{x} \quad\left(\because \mathrm{u}_{\mathrm{CD}}=\mathrm{v}_{\mathrm{BC}}\right)\)
\(0=6 g x \sin \theta-3 \mu g x \cos \theta\)
\(6 g x \sin \theta=3 \mu g x \cos \theta\)
\(\mu=\frac{6 g x \sin \theta}{3 g x \cos \theta}\)
\(\mu=2 \tan \theta\)

AP EAMCET (22.04.2019) Shift-II

Laws of Motion

146163 A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontal by with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(\mathrm{g}=10 \mathrm{~ms}^{-2}\) )

1 \(\frac{2}{9}\)

2 \(\frac{5}{12}\)

3 \(\frac{1}{5}\)

4 \(\frac{2}{5}\)

AP EAMCET (21.04.2019) Shift-I

Laws of Motion

146164 The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is \(60^{\circ}\), then the coefficient of friction is

1 \(\frac{1}{3}\)

2 \(\frac{1}{\sqrt{2}}\)

3 \(\frac{1}{\sqrt{3}}\)

4 \(\frac{1}{2}\)

AP EAMCET (20.04.2019) Shift-II

Laws of Motion

146165 \(\quad\) The maximum value of the applied force \(F\) such that the block as shown in the arrangement does not move is
(Acceleration due to gravity, \(g=10 \mathbf{m s}^{-2}\) )

1 \(20 \mathrm{~N}\)

2 \(15 \mathrm{~N}\)

3 \(25 \mathrm{~N}\)

4 \(10 \mathrm{~N}\)

AP EAMCET (20.04.2019) Shift-1

Laws of Motion

146166 A body of mass \(2 \mathrm{~kg}\) is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 \(\mathrm{N}\), then the frictional force acting on the body will be \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\)

1 \(8 \mathrm{~N}\)

2 \(10 \mathrm{~N}\)

3 \(20 \mathrm{~N}\)

4 \(2.5 \mathrm{~N}\)

AIIMS-26.05.2018(M)

Explanation:

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Question Bank Series

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