Physics

Friction and Inclined Planes

58 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

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
A Correct answer
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 surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

A metal plane having an area of $0.04\ m^{2}$ is placed on a horizontal wooden surface. Oil of coefficient of viscosity $2\ N/ s/m^{2}$ is introduced between the plate and the surface till the thickness of the oil layer is $0.5$ in. The horizontal force needed to drag the plate along the surface with a velocity of $5\ cm/s$ is  

  1. $80\ N$
  2. $8\ N$
  3. $60\ N$
  4. $6\ N$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 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
Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

A block of mass $100\ g$ slides on a rough horizontal surface. If the speed of the block decreases from $10\ m/s^{-1}$ to $5\ m/s^{-1}$, the thermal energy developed in the process is:

  1. $3.75\ J$
  2. $37.5\ J$
  3. $0.375\ J$
  4. $0.75\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given:
The mass of the block is $100\ g$
The initial speed of the block is $10\ m/s$
The final speed of the block is $5\ m/s$

The thermal energy developed in the process is due to the lowering of the speed i.e. the reduction in the kinetic energy of the block.

Thermal energy = loss in kinetic energy
$=\dfrac{1}{2}m(v^2 _1-v^2 _2)$

$=\dfrac{1}{2}100\times 10^{-3}(10^2-5^2)$

$=3.75J$
Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A highly rigid cubical block A of small mass M and side L is fixed rigidly on another cubical block B of the same dimensions and of low modulus of rigidity $\eta $ such that the lower face of A completely covers the upper face of B.  The lower face of B is rigidly held on horizontal surface.  A small force is applied perpendicular to the side faces of A.  After the force is withdrawn, block A executes small oscillations the time period of which is given by 

  1. $2\pi \sqrt{M\eta L}$
  2. $2\pi \sqrt{\frac{M-\eta }{L}}$
  3. $2\pi \sqrt{\frac{M-L}{\eta }}$
  4. $2\pi \sqrt{\frac{M-N}{\eta L}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice power work and power work, energy and power physics energy and its forms

A rectangular block of dimensions $6m\times 4m\times 2m$ and of density $1.5\ gm/c.c$ is lying on horizontal ground with the face of large area in contact with the ground. The work done in arranging it which its smallest area in contact with a ground is, $(g=10ms^{-1})$

  1. $2880\ kJ$
  2. $1440\ kJ$
  3. $3800\ kJ$
  4. $720\ kJ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

m=v*density

m=$6\times 4\times 2\times 1.5\times 10^3=72\times 10^3$kg
when face with large area is in contact with the ground its height is 2m
Centre of mass is at a height of 1m

when face with small area is in contact with the ground its height is 6m
Centre of mass is at a height of 3m
W=$72\times 10^4(3-1)=1440 kJ$

Multiple choice power work and power work, energy and power physics energy and its forms

A small body of mass $m$ is located on a horizontal plane. The body acquires a horizontal velocity ${v} _{0}$. Find mean power developed by the frictional force, during the whole time of its motion. Coefficient of friction is $\mu$

  1. $\cfrac { -\mu mg{ v } _{ 0 } }{ 3 } $
  2. $\cfrac { -\mu mg{ v } _{ 0 } }{ 2 } $
  3. $\cfrac { -\mu mg{ v } _{ 0 } }{ 5 } $
  4. $\cfrac { -\mu mg{ v } _{ 0 } }{ 6 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The body decelerates due to friction: a = -mu * g. Final velocity is 0. Time taken t = v0 / (mu * g). Displacement s = v0^2 / (2 * mu * g). Work done = -mu * m * g * s = -m * v0^2 / 2. Mean power = Work / t = (-m * v0^2 / 2) / (v0 / mu * g) = -mu * m * g * v0 / 2.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two blocks m and m each of mass 3kg is connected with spring of constant 50 N/m. The coefficient of friction between m and ground is 0.4. The maximum amplitude of m during its oscillation, so that m does not move, is 

  1. 24 cm

  2. 12 cm

  3. 2.4 cm

  4. 6 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the block not to move, the maximum spring force must be less than or equal to the limiting friction. k*A <= mu*m*g. 50 * A <= 0.4 * 3 * 10. 50 * A <= 12. A <= 12/50 = 0.24 m = 24 cm.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

The backside of a truck is open and a box of 40kg is placed 5m away from the rear end.The coefficient of friction of the box with the surface of the truck is 0.15.The truck starts from rest with $2m/s^2$ acceleration.Calculate the distance covered by the truck when the box falls off

  1. 20m

  2. 30m

  3. 40m

  4. 50m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The box experiences a pseudo-force F = m*a = 40 * 2 = 80N. The friction force is f = mu * m * g = 0.15 * 40 * 10 = 60N. The net force on the box is F_net = F - f = 80 - 60 = 20N. Acceleration of the box relative to the truck is a_rel = F_net / m = 20 / 40 = 0.5 m/s^2. Time to fall off: s = 0.5 * a_rel * t^2 => 5 = 0.5 * 0.5 * t^2 => 5 = 0.25 * t^2 => t^2 = 20 => t = sqrt(20). Distance covered by truck: S = 0.5 * a_truck * t^2 = 0.5 * 2 * 20 = 20m.