Physics

Friction and Inclined Planes

63 Questions

Friction and inclined plane problems focus on calculating forces, coefficients of static and kinetic friction, and motion on rough surfaces. Questions cover blocks on horizontal tables, angled planes, and simple harmonic motion systems. These physics concepts are strictly evaluated in engineering entrance and civil services tests.

Static friction coefficientKinetic friction forceInclined plane blocksDamped vibrations systemHorizontal table motion

Friction and Inclined Planes Questions

Multiple choice friction laws of motion physics

The coefficient of friction between two surfaces is 0.2. The angle of friction is 

  1. sin$^{-1}$(0.2)
  2. cos $^{-1}$(0.2)
  3. tan$^{-1}$(0.1)
  4. cot$^{-1}$(5)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The correct option is D

We have,

 The coefficient of friction is $0.2$

Since we know that,

The coefficient of friction $= tan \theta$  Where $\theta $  is the angle of friction.

$\dfrac{1}{5}=tan \theta$

$\theta=tan^{-1}\dfrac{1}{5}$

$=cot^{-1}5$
Multiple choice friction laws of motion physics

If angle of repose is ${30}^{o}$, then coefficient of friction will be

  1. $1$
  2. $15$
  3. $\cfrac { 1 }{ \sqrt { 3 } } $
  4. $\cfrac { \sqrt { 3 } }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\mu =tan\theta \ =tan30=\frac { 1 }{ \sqrt { 3 }  } \ $


Multiple choice friction laws of motion physics

The coefficient of friction between a chain & a table is m. If a chain is placed on a horizontal table so that a part of it is hanging from one end, the minimum fraction of length of the chain that can be on the table, so that the chain may not slip off is

  1. $\frac{\mu}{\mu + 1}$
  2. $\frac{1}{\mu + 1}$

  3. 1

  4. Zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the chain not to slip, the friction force on the table portion must equal the weight of the hanging portion. If L is the total length and x is the length on the table, then mu * (x/L) * m * g = ((L-x)/L) * m * g. Solving for the fraction x/L gives mu/(mu+1).

Multiple choice friction laws of motion physics

A piece of wood of mass $150$ g rests on an inclined plane. The co-efficient of friction between the surfaces in contact is $0.3$. To what maximum extent the plane may be inclined without allowing the piece to clip down?

  1. $26.7^o$.
  2. $16.7^o$.
  3. $36.7^o$.
  4. $46.7^o$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum angle of inclination before sliding occurs is given by tan(theta) = mu. With mu = 0.3, theta = arctan(0.3), which is approximately 16.7 degrees.

Multiple choice friction laws of motion physics

The upper half of an inclined plane with inclination $\alpha $ is perfectly smooth while the lower half is rough, a body starts from rest at the top of the inclined and comes to rest again at the bottom of it. The coefficient of friction for the lower half of the incline is:

  1. $\frac { 1 }{ 2 } tan\alpha $
  2. $2sin\alpha $
  3. $cot\alpha $
  4. $2tan\alpha $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using work-energy principles or equations of motion, the work done by gravity down the entire incline must equal the work done by friction along the rough lower half. Equating these components leads to the coefficient of friction being equal to 2 * tan(alpha).

Multiple choice friction laws of motion physics

A block is kept on a horizontal table. The table is undergoing simple harmonic motion of frequency $3\, Hz$ in a horizontal plane. The coefficient of static friction between the block and the table surface is $0.72$. Find the maximum amplitude of the table at which the block does not slip on the surface $(g = 10\, ms^{-2})$

  1. $0.01m$
  2. $0.02m$
  3. $0.03m$
  4. $0.04m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum acceleration of the table is a_max = omega^2 * A = (2*pi*f)^2 * A. For the block not to slip, a_max <= mu*g. Substituting f=3, mu=0.72, and g=10, we get (2*pi*3)^2 * A <= 7.2. Solving for A gives 7.2 / (36 * pi^2) approx 0.02m.

Multiple choice friction laws of motion physics

A block of mass $2\ kg$ rests on a rough inclined plane making an angle of ${30}^{o}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

  1. $9.8\ N$
  2. $0.7\times 9.8\times \sqrt { 3 } N$
  3. $9.8\times \sqrt { 3 } N$
  4. $0.7\times 9.8\ N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force applied on the body that is on the inclined plane is given as,

$F = mg\sin \theta $

$F = 2 \times 9.8 \times \sin 30^\circ $

$ = 9.8\;{\rm{N}}$

The limiting friction force between the block and the inclined plane is given as,

$f = \mu mg\cos \theta $

$f = 0.7 \times 2 \times 9.8\cos 30^\circ $

$ = 11.88\;{\rm{N}}$

Since the limiting friction force is greater than the force that tends to slide the body.

Thus, the body will be at rest and the force of friction on the block is $9.8\;{\rm{N}}$.

Multiple choice friction laws of motion physics

An object is placed on the surface of a smooth inclined plane of inclination $\theta$. It takes time $t$ to reach the bottom. If the same object is allowed to slide down a rough inclined plane of same inclination $\theta $, it takes times nth to reach the bottom where $n$ number greater than $1$. The coefficient of friction $\mu$ is given by:

  1. $\mu =\tan { \theta \left( 1-1/{ n }^{ 2 } \right) } $
  2. $\mu =\cot { \theta \left( 1-1/{ n }^{ 2 } \right) } $
  3. ${ \mu =\tan { \theta \left( 1-1/{ n }^{ 2 } \right) } }^{ 1/2 }$
  4. ${ \mu =\cot { \theta \left( 1-1/{ n }^{ 2 } \right) } }^{ 1/2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac { { V } _{ a } }{ n } =\sqrt { 2Lg\left( \sin\theta -\mu \cos\theta  \right)  } \quad \longrightarrow \left( 1 \right) $

${ V } _{ a }=\sqrt { 2Lg\sin\theta  } \quad \longrightarrow \left( 2 \right) $
By putting equ(2) in eq(1) we get
$\dfrac { \sqrt { 2Lg\sin\theta  }  }{ n } =\sqrt { 2Lg\left( \sin\theta -\mu \cos\theta  \right)  } $
$\dfrac { 2Lg\sin\theta  }{ { n }^{ 2 } } =2Lg\left( \sin\theta -\mu \cos\theta  \right) $
$\dfrac { \sin\theta  }{ { n }^{ 2 } } =\sin\theta -\mu \cos\theta $
$\mu \cos\theta =\sin\theta -\dfrac { \sin\theta  }{ { n }^{ 2 } } $
$\mu \cos\theta =\sin\theta \left( 1-\dfrac { 1 }{ { n }^{ 2 } }  \right) $
$\mu =\tan\theta \left( 1-\dfrac { 1 }{ { n }^{ 2 } }  \right) $

Multiple choice friction laws of motion physics

A body of  weight 20 N is on a horizontal surface, minimum force applied to pull it when applied force makes an angle $60^0$ with horizontal (angle of friction a = $30^0$) is:

  1. 20 N
  2. 20 $\sqrt{3}$ N
  3. $\dfrac{20}{\sqrt{3}}$ N

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For minimum force, we will use limiting friction  i.e.
$\Rightarrow \tan\phi=\mu _{s}$
and ATQ, $\phi=30$
$\Rightarrow \tan 30=\mu _{s}$
$\Rightarrow \mu _{s}=\cfrac{1}{\sqrt 3}$
Minimum force can be calculated as $=\mu _{s}mg$
ATQ, $mg=20N, \mu=\cfrac{1}{\sqrt 3}$
$\Rightarrow F _{min}=\cfrac{20}{\sqrt 3}N$

Multiple choice force in shm oscillations oscillation and waves physics

A coin is placed on a horizontal platform, which undergoes horizontal simple harmonic motion about a mean position $O$.The coin does not slip on the platform. The force of friction acting on the coin is $F$.

  1. $F$ is always directed towards $O$
  2. $F$ is directed towards $O$ when the coin is moving away from $O$, and away from $O$ when the coin moves towards $O$
  3. $F=0$ when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.
  4. $F$ is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In horizontal SHM, the platform accelerates towards the mean position O. To prevent slipping, the static friction force must provide this acceleration, meaning the friction force must always be directed towards O.

Multiple choice physics friction increasing and reducing friction advantages and disadvantages of friction merits and demerits of friction and methods to reduce it

Correct air pressure in our vehicle tyres help to reduce:

  1. static friction

  2. sliding friction

  3. rolling friction

  4. all of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Correct air pressure helps in reducing deformation of the tyre and also provide an air cushion to reduce rolling friction. 


Option C is correct.

Multiple choice physics friction increasing and reducing friction advantages and disadvantages of friction merits and demerits of friction and methods to reduce it

Why are efforts made to reduce friction?

  1. Energy is lost in overcoming friction.

  2. Overcoming friction is a time consuming process.

  3. Both A and B

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A lot of the applied energy is lost in overcoming friction so as to put a body in motion.Hence efforts are made to reduce friction.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

In the game of cricket, the stumps falls when the ball strikes them. This is an example of

  1. Contact force

  2. Non contact force

  3. Displacement force

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
When a speeding ball strakees stumps, they came in contact with each other, due to that a contact force greater between the two.
This face pushes the $3$ function in the direction of motion of the ball resulting it to fall off. 
option $A$ is correct






Multiple choice physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

A uniform dice of mass $10kg$ radius $1m$ is placed on a rought horizontal surface. The coefficient of friction between the disc and the surface is $0.2$. A horizontal time varying force is applied on the centre of the disc whose variation with time is shown in graph.
List-I                                                         List-IIDisc rolls without slipping                   at $t=7s$Disc rolls with slipping                       at $t=3s$  Disc starts slipping at                         at $t=4s$Friction force is $10N$ at              None

  1. $A-p,q;B-p;C-r;Dq$
  2. $A-p,r;B-s;C-s,p;D-q$
  3. $A-q,r;B-p;C-s;D-q$
  4. $A-p,q,r;B-q;r;C-s;p;D-p,q,r,s$
Reveal answer Fill a bubble to check yourself
C Correct answer