30 Friction MCQs with Answers for High School Physics

A block of mass m is placed on a rough horizontal surface. The coefficient of static friction is \(\mu_s\). A horizontal force F is gradually increased from zero. What is the friction force when \(F < \mu_s mg\)?

1. \(\mu_s mg\)
2. \(\dfrac{F}{2}\)
3. F
4. 0

Explanation

Static friction is a self-adjusting force.

As long as the block remains at rest,

\[
f_s=F
\]

provided

\[
F\le \mu_s mg
\]

The friction force only reaches its maximum value at impending motion.

A block rests on a rough incline of angle \(\theta\). The coefficient of static friction is \(\mu_s\). Which condition ensures that the block remains at rest?

1. \(\tan\theta>\mu_s\)
2. \(\tan\theta\le\mu_s\)
3. \(\sin\theta\le\mu_s\)
4. \(\cos\theta\le\mu_s\)

Explanation

For equilibrium,

\[
mg\sin\theta\le\mu_s mg\cos\theta
\]

Thus,

\[
\tan\theta\le\mu_s
\]

A block of mass 4 kg is placed on a rough horizontal surface with \(\mu_s=0.5\). What is the minimum horizontal force required to start motion? Take \(g=10\,m/s^2\).

1. 10 N
2. 15 N
3. 20 N
4. 25 N

Explanation

Maximum static friction:

\[
f_{max}=\mu_s N
\]

\[
=0.5\times4\times10
\]

\[
=20N
\]

The applied force must exceed 20 N to start motion.

Two blocks of masses m and 2m are stacked. The coefficient of static friction between them is \(\mu\). The lower block is pulled horizontally. What is the maximum acceleration of the system so that the upper block does not slip?

1. \(\dfrac{\mu g}{2}\)
2. \(\mu g\)
3. \(2\mu g\)
4. \(3\mu g\)

Explanation

For the top block,

\[
f=ma
\]

Maximum available friction:

\[
f_{max}=\mu mg
\]

Thus,

\[
ma\le\mu mg
\]

\[
a\le\mu g
\]

A block is placed on a rough horizontal surface. The coefficient of kinetic friction is 0.2. It is given an initial speed of 10 m/s. How far does it travel before stopping? Take \(g=10\,m/s^2\).

1. 15 m
2. 20 m
3. 22.5 m
4. 25 m

Explanation

Retardation:

\[
a=\mu_k g
\]

\[
=0.2\times10
\]

\[
=2\,m/s^2
\]

Using

\[
v^2=u^2-2as
\]

\[
0=100-4s
\]

\[
s=25m
\]

A ladder leans against a smooth wall. The ground is rough. Which force prevents the ladder from slipping?

1. Normal force from wall
2. Weight of ladder
3. Friction at the ground
4. Tension in ladder

Explanation

Since the wall is smooth, it cannot provide friction.

Only the friction force at the ground balances the horizontal reaction from the wall.

A block of mass m is pressed against a vertical wall by a horizontal force P. The coefficient of static friction between the block and wall is \(\mu\). What is the minimum value of P required to prevent the block from falling?

1. \(\dfrac{mg}{2\mu}\)
2. \(\dfrac{mg}{\mu}\)
3. \(\mu mg\)
4. \(\dfrac{\mu mg}{2}\)

Explanation

Normal reaction:

\[
N=P
\]

Maximum friction:

\[
f_{max}=\mu P
\]

For equilibrium,

\[
\mu P\ge mg
\]

Thus,

\[
P\ge\dfrac{mg}{\mu}
\]

A block rests on a rough incline. The angle of inclination is gradually increased. At the instant the block is about to slide, which statement is correct?

1. \(f<\mu_sN\)
2. \(f=\mu_sN\)
3. \(f>\mu_sN\)
4. \(f=0\)

Explanation

At limiting equilibrium,

\[
f_{lim}=\mu_sN
\]

This is the maximum possible static friction.

A block of mass m lies on a rough horizontal surface. A force F acts at an angle θ above the horizontal. How does increasing θ affect the limiting friction?

1. It increases
2. It decreases
3. It remains constant
4. It first increases then decreases

Explanation

Normal force becomes

\[
N=mg-F\sin\theta
\]

Limiting friction:

\[
f_{max}=\mu_sN
\]

As θ increases, \(N\) decreases, therefore limiting friction also decreases.

A block of mass m is placed on a wedge accelerating horizontally with acceleration a. The coefficient of static friction is sufficient to prevent slipping. In the wedge frame, which force is responsible for balancing the pseudo force component along the incline?

1. Normal reaction only
2. Gravity only
3. Static friction
4. Kinetic friction

Explanation

In the non-inertial frame of the wedge, the pseudo force has a component along the incline.

For the block to remain at rest relative to the wedge, static friction adjusts itself to balance the net force along the plane.

A wedge of angle \(\theta\) is accelerated horizontally with acceleration \(a\). A block rests on the wedge without slipping. If the coefficient of friction is just sufficient, the friction force acts:

1. Always up the incline
2. Always down the incline
3. Either up or down the incline depending on \(a\)
4. Perpendicular to the incline

Explanation

In the wedge frame, a pseudo force acts opposite to the wedge acceleration.

The component of the effective force along the incline may point either upward or downward depending on the magnitude of \(a\).

Therefore, static friction adjusts its direction accordingly.

A block of mass m rests on a rough wedge of angle \(\theta\). The coefficient of static friction is \(\mu\). What is the maximum angle of inclination for equilibrium?

1. \(\theta=\sin^{-1}\mu\)
2. \(\theta=\cos^{-1}\mu\)
3. \(\theta=\tan^{-1}\mu\)
4. \(\theta=\cot^{-1}\mu\)

Explanation

At limiting equilibrium,

\[
mg\sin\theta=\mu mg\cos\theta
\]

Therefore,

\[
\tan\theta=\mu
\]

Thus,

\[
\theta_{max}=\tan^{-1}\mu
\]

Two blocks of masses m and 3m are connected by a light string over a smooth pulley. The block of mass m rests on a rough horizontal surface with coefficient of kinetic friction \(\mu\). What is the acceleration of the system?

1. \(\dfrac{g(3-\mu)}{3}\)
2. \(\dfrac{g(3-\mu)}{4}\)
3. \(\dfrac{g(1+\mu)}{4}\)
4. \(\dfrac{g(3+\mu)}{4}\)

Explanation

For the block on the table,

\[
T-\mu mg=ma
\]

For the hanging block,

\[
3mg-T=3ma
\]

Adding,

\[
3mg-\mu mg=4ma
\]

\[
a=\dfrac{g(3-\mu)}{4}
\]

A block is projected up a rough incline with speed u. The coefficient of kinetic friction is \(\mu\). What is the retardation while moving upward?

1. \(g(\sin\theta-\mu\cos\theta)\)
2. \(g(\sin\theta+\mu\cos\theta)\)
3. \(\mu g\cos\theta\)
4. \(g\sin\theta\)

Explanation

Both gravity and friction act down the plane.

Hence,

\[
a=g\sin\theta+\mu g\cos\theta
\]

This acts opposite to the motion.

A block slides down a rough incline with coefficient of kinetic friction \(\mu\). Its acceleration is:

1. \(g(\sin\theta+\mu\cos\theta)\)
2. \(g(\sin\theta-\mu\cos\theta)\)
3. \(\mu g\sin\theta\)
4. \(g\cos\theta\)

Explanation

Along the incline,

\[
mg\sin\theta-f_k=ma
\]

with

\[
f_k=\mu mg\cos\theta
\]

Thus,

\[
a=g(\sin\theta-\mu\cos\theta)
\]

A block of mass m rests on another block of mass M. The lower block is pulled by a force F. If there is no slipping, the friction force on the upper block is:

1. Always \(\mu mg\)
2. \(\dfrac{F}{2}\)
3. \(ma\)
4. Zero

Explanation

The only horizontal force acting on the upper block is friction.

Therefore,

\[
f=ma
\]

where a is the common acceleration of the system.

A block rests against a rough vertical wall. The coefficient of friction is \(\mu\). If the horizontal force pressing the block is doubled, the maximum possible friction becomes:

1. Unchanged
2. Twice its previous value
3. Half its previous value
4. Four times its previous value

Explanation

Since

\[
f_{max}=\mu N
\]

and

\[
N=P
\]

doubling P doubles N.

Hence the limiting friction also doubles.

A uniform ladder rests against a rough wall and a rough floor. If friction at the wall is neglected, which force provides the torque needed to prevent slipping?

1. Weight only
2. Normal reaction at floor only
3. Friction force at the floor
4. Normal reaction at wall only

Explanation

The wall exerts only a horizontal reaction.

The balancing torque necessary for equilibrium comes from the friction force acting at the floor.

A block of mass m is placed on a rough horizontal surface. A horizontal force \(F=2\mu mg\) acts on it. The coefficient of kinetic friction is \(\mu\). What is the acceleration of the block?

1. \(\mu g\)
2. \(g\mu\)
3. \(2\mu g\)
4. \(3\mu g\)

Explanation

Kinetic friction:

\[
f_k=\mu mg
\]

Net force:

\[
F_{net}=2\mu mg-\mu mg
\]

\[
=\mu mg
\]

Therefore,

\[
a=\dfrac{\mu mg}{m}
\]

\[
a=\mu g
\]

A block rests on a rough incline inside an elevator accelerating upward with acceleration a. The maximum static friction becomes:

1. \(\mu mg\)
2. \(\mu m(g-a)\)
3. \(\mu m(g+a)\)
4. \(\mu ma\)

Explanation

In the accelerating elevator,

\[
N=m(g+a)\cos\theta
\]

For a horizontal floor, simply

\[
N=m(g+a)
\]

Hence,

\[
f_{max}=\mu N
\]

\[
f_{max}=\mu m(g+a)
\]

This is a classic non-inertial frame friction result.

A block of mass m is placed on a rough wedge of angle \(\theta\). The wedge is accelerated horizontally such that the block remains at rest relative to the wedge without any friction. The required acceleration of the wedge is:

1. \(g\sin\theta\)
2. \(g\cos\theta\)
3. \(g\tan\theta\)
4. \(\dfrac{g}{\tan\theta}\)

Explanation

In the wedge frame, the pseudo force \(ma\) acts horizontally.

For no tendency of motion along the incline,

\[
ma\cos\theta=mg\sin\theta
\]

\[
a=g\tan\theta
\]

At this acceleration, friction is not required.

A block is placed on a rough horizontal turntable rotating with angular speed \(\omega\). The coefficient of static friction is \(\mu\). What is the maximum distance from the center at which the block can remain at rest relative to the turntable?

1. \(\dfrac{\mu g}{\omega}\)
2. \(\dfrac{\mu g}{\omega^2}\)
3. \(\dfrac{\mu g}{\omega^2}\)
4. \(\dfrac{\omega^2}{\mu g}\)

Explanation

Static friction provides the centripetal force.

\[
m\omega^2r\le\mu mg
\]

Therefore,

\[
r\le\dfrac{\mu g}{\omega^2}
\]

A plank of mass M rests on a frictionless surface. A block of mass m is placed on top of it. The coefficient of friction between them is \(\mu\). What is the maximum horizontal force that can be applied to the plank so that the block does not slip?

1. \(\mu mg\)
2. \(\mu g(M+m)\)
3. \(\mu gM\)
4. \(\mu g(M-m)\)

Explanation

Common acceleration:

\[
a=\dfrac{F}{M+m}
\]

For no slipping,

\[
ma\le\mu mg
\]

\[
a\le\mu g
\]

Thus,

\[
\dfrac{F}{M+m}\le\mu g
\]

\[
F_{max}=\mu g(M+m)
\]

A block slides down a rough incline. The coefficient of kinetic friction is \(\mu\). For what value of \(\theta\) will the acceleration be maximum?

1. \(\tan\theta=\mu\)
2. \(\theta=45^\circ\)
3. \(\theta=90^\circ\)
4. \(\theta=0^\circ\)

Explanation

Acceleration is

\[
a=g(\sin\theta-\mu\cos\theta)
\]

Differentiating,

\[
\dfrac{da}{d\theta}
=
g(\cos\theta+\mu\sin\theta)
\]

The acceleration increases continuously as \(\theta\) increases.

Maximum value occurs at

\[
\theta=90^\circ
\]

where

\[
a=g
\]

A block of mass m rests on a rough horizontal surface. A force F is applied at an angle \(\theta\) below the horizontal. How does the limiting friction compare with the case when the same force is applied horizontally?

1. It decreases
2. It remains unchanged
3. It increases
4. It becomes zero

Explanation

The downward component increases the normal reaction.

\[
N=mg+F\sin\theta
\]

Therefore,

\[
f_{max}=\mu N
\]

which is larger than before.

A uniform rod rests against a rough wall and a rough floor. If the coefficient of friction is the same at both contacts, which quantity must be satisfied for equilibrium?

1. Only force balance
2. Only torque balance
3. Both force and torque balance simultaneously
4. Only friction balance

Explanation

Ladder and rod problems are rigid-body equilibrium problems.

The conditions are:

\[
\sum F_x=0
\]

\[
\sum F_y=0
\]

\[
\sum\tau=0
\]

All must hold simultaneously.

A block of mass m is placed on another block of mass M. The coefficient of friction between them is \(\mu\). If the lower block is accelerated with \(a>\mu g\), what happens?

1. The blocks move together
2. The upper block slips backward relative to the lower block
3. The upper block slips forward relative to the lower block
4. Both remain at rest

Explanation

Maximum acceleration that friction can provide is

\[
a_{max}=\mu g
\]

If

\[
a>\mu g
\]

friction cannot accelerate the upper block sufficiently.

Therefore, it slips backward relative to the lower block.

A block moves on a rough horizontal surface. The coefficient of kinetic friction depends on speed according to \(\mu=k/v\). What is the magnitude of the retardation?

1. \(kg\)
2. \(\dfrac{kg}{v}\)
3. \(\dfrac{v}{kg}\)
4. \(\dfrac{kg}{v^2}\)

Explanation

Friction force:

\[
f=\mu N
\]

\[
=\dfrac{k}{v}mg
\]

Therefore,

\[
a=\dfrac{f}{m}
\]

\[
a=\dfrac{kg}{v}
\]

A block is placed on a rough incline. The angle is such that \(\tan\theta=\mu\). The block is given a small downward push. Which statement is correct?

1. It remains at rest forever
2. It moves with constant acceleration \(g\sin\theta-\mu g\cos\theta=0\) initially, but kinetic friction determines subsequent motion
3. It accelerates with g
4. It moves upward

Explanation

At

\[
\tan\theta=\mu
\]

the system is at limiting equilibrium.

Once motion starts, kinetic friction becomes relevant.

If \(\mu_k<\mu_s\), the block accelerates downward. This is a classic conceptual trap in advanced friction problems.

A block of mass m is kept on a rough rotating disc. The angular speed is gradually increased. What is the first quantity that reaches its limiting value before slipping occurs?

1. Normal reaction
2. Weight
3. Static friction
4. Kinetic friction

Explanation

Static friction supplies the required centripetal force.

As \(\omega\) increases,

\[
f_s=m\omega^2r
\]

increases continuously.

Slipping begins when

\[
f_s=\mu_sN
\]

Thus static friction reaches its limiting value first.