Common forces in mechanics MCQ Quiz - Objective Question with Answer for Common forces in mechanics - Download Free PDF

Last updated on May 11, 2025

Latest Common forces in mechanics MCQ Objective Questions

Common forces in mechanics Question 1:

A block of mass m₁ = 1 kg and another block of mass m₂ = 2 kg are placed together on an inclined plane with angle of inclination θ . Various values of θ are given in Column I. The coefficient of friction between the block m₁ and the plane is always zero. The coefficient of static and dynamic friction between the block m₂ and the plane are equal to μ = 0.3. In Column II, expressions for the friction on block m₂ are given. Match the correct expression of the friction in Column II with the angles given in Column I. The acceleration due to gravity is denoted by g. [Given: tan(5.5°) ≈ 0.1, tan(11.5°) ≈ 0.2, tan(16.5°) ≈ 0.3]

Column I Column II
(P) θ = 5° (1) m₂g sin θ
(Q) θ = 10° (2) (m₁ + m₂)g sin θ
(R) θ = 15° (3) μm₂g cos θ
(S) θ = 20° (4) μ(m₁ + m₂)g cos θ

  1. P → 1, Q → 2, R → 2, S → 3
  2. P → 2, Q → 2, R → 3, S → 3
  3. P → 2, Q → 1, R → 3, S → 1
  4. P → 1, Q → 1, R → 2, S → 2

Answer (Detailed Solution Below)

Option 2 : P → 2, Q → 2, R → 3, S → 3

Common forces in mechanics Question 1 Detailed Solution

Calculation:

The forces on block m₁ are: its weight m₁g, normal reaction from the inclined plane N₁, and the reaction from block m₂ as R.

The forces on block m₂ are: m₂g, N₂, R, and the frictional force f.

If θ is slowly increased, f starts increasing and reaches its maximum value f = μN₂ at θ = θr (angle of repose).

From Newton’s second law:

f = μN₂

R = m₁g sin θ

N₁ = m₁g cos θ

f = R + m₂g sin θ

N₂ = m₂g cos θ

qImage681d9f91fee88e53e8a763dd

Solving these, we get the angle of repose:

θr = tan−1[(μ × m₂) / (m₁ + m₂)] = tan−1[(0.3 × 2) / (1 + 2)] = tan−1(0.2) = 11.5°

So:

For θ = 5° and θ = 10°, blocks are at rest with frictional force:

f = R + m₂g sin θ = (m₁ + m₂)g sin θ

For θ = 15° and θ = 20°, blocks are moving with frictional force:

f = μN₂ = μm₂g cos θ

Common forces in mechanics Question 2:

A body of weight \(64\, N\) is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and kinetic friction are \(0.6\) and \(0.4\) respectively, then the acceleration of the block will be (Acceleration due to gravity \(g\))

  1. \(\dfrac{g}{6.4}\)
  2. \(0.64\, g\)
  3. \(\dfrac{g}{32}\)
  4. 1.96
  5. \(\dfrac{g}{3.2}\)

Answer (Detailed Solution Below)

Option 4 : 1.96

Common forces in mechanics Question 2 Detailed Solution

Calculation:
According to Newton's third law, the force just enough to start the body moving must be equal to the force of static friction. And, according to Amontons' laws of friction,

F = Ffriction, static

Ffriction, static = μstatic N

N - mg = 0 (y-axis projection)

N = mg

Ffriction, static = μstatic mg

F = μstatic mg

qImage671b295cd548c9faf3ee46f1

According to Newton's second law,

F - Ffriction, dynamic = ma (x-axis projection)

Ffriction, dynamic = μdynamic N

N - mg = 0 (y-axis projection)

N = mg

Ffriction, dynamic = μdynamic mg

F - μdynamic mg = ma

μstatic mg - μdynamic mg = ma

a = g (μstatic - μdynamic)

a = 9.8 (0.6 - 0.4)

a = 1.96 m/s2

Common forces in mechanics Question 3:

A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μs' then the magnitude of negative lift FL acting downwards on the car is: (Assume forces on the four tyres are identical and g = acceleration due to gravity)

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  1. \(m\left( {\frac{{{v^2}}}{{{\mu _s}R}} + g} \right)\)
  2. \(m\left( {g-\frac{{{v^2}}}{{{\mu _s}R}} } \right)\)
  3. \(m\left( {\frac{{{v^2}}}{{{\mu _s}R}} - g} \right)\)
  4. \(-m\left( {g+\frac{{{v^2}}}{{{\mu _s}R}} } \right)\)
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : \(m\left( {\frac{{{v^2}}}{{{\mu _s}R}} - g} \right)\)

Common forces in mechanics Question 3 Detailed Solution

CONCEPT:

→The normal force in the circular track of radius R is written as;

 \(N = \frac{mv^2}{\mu R}\)

Here N is normal, m is the mass, R is the radius and v is the velocity.

CALCULATION:

car of mass m is traveling on a flat track in a circular arc of radius R with a speed v, the free body diagram is shown below;

qImage23951

From the diagram we have;

\(mg +F_L = N\)   ----(1)

Ad we know, \(N = \frac{mv^2}{\mu R}\)

By putting the values of N in equation (1) we have;

\(mg +F_L = \frac{mv^2}{\mu R}\)

⇒ \(F_L = \frac{mv^2}{\mu R}-mg\)

⇒ \(F_L = m(\frac{v^2}{\mu R}-g)\)

Hence, option 3) is the correct answer.

Common forces in mechanics Question 4:

A block of mass 20 kg slides on a rough horizontal plane surface. Let the speed of the block at a particular instant is 10 m/s. It comes to rest after travelling a distance of 20 m. Which one of the following could be the magnitude of the frictional force?

  1. 10 N
  2. 20 N
  3. 40 N
  4. 50 N

Answer (Detailed Solution Below)

Option 4 : 50 N

Common forces in mechanics Question 4 Detailed Solution

Solution:

The block comes to rest after traveling a distance of 20 m. We can use the work-energy principle to find the frictional force. The work done by the frictional force is equal to the change in kinetic energy of the block.

The initial kinetic energy (K.E.) of the block is given by:

K.E. = (1/2) × m × v² = (1/2) × 20 × (10)² = 1000 J

The work done by the frictional force is:

W = F × d

Where F is the frictional force and d is the distance traveled (20 m). Since the block comes to rest, all the kinetic energy is used up by the frictional force. Therefore:

1000 J = F × 20

Solving for F:

F = 1000 / 20 = 50 N

Hence, the magnitude of the frictional force is 50 N (Option 4).

Common forces in mechanics Question 5:

A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is 30°. The coefficients of static and kinetic frictions between the block and the inclined plane are 0.6 and 0.5 respectively Then the magnitude of the acceleration of the block is [Take g = 10 ms-2

qImage670641ef16972ef286c25a3a

  1. 2 ms-2
  2. zero 
  3. 0.196 ms-2
  4. 0.67 ms-2
  5. 0.96 ms-2

Answer (Detailed Solution Below)

Option 2 : zero 

Common forces in mechanics Question 5 Detailed Solution

Given:

A block of mass m is placed on a rough inclined plane at an angle of 30° with the horizontal. Coefficients of static friction (μs) and kinetic friction (μk) are 0.6 and 0.5, respectively. The gravitational acceleration g = 10 m/s². We need to calculate the acceleration of the block.

Concept:

  • The forces acting on the block are:
    • Gravitational force along the incline: mgsinθ.
    • Normal reaction force perpendicular to the incline: mgcosθ.
    • Frictional force opposing motion:
      • Static friction: fs ≤ μsmgcosθ.
      • Kinetic friction: fk = μkmgcosθ.
  • The block will remain stationary if the static friction is sufficient to balance the gravitational force along the incline:

    fs ≥ mgsinθ.

  • If mgsinθ > μsmgcosθ, the block will move, and the net acceleration can be calculated as:

    a = gsinθ - μkgcosθ.

Calculation:

Calculate mgsinθ and μsmgcosθ:

mgsinθ = m × 10 × sin30° = m × 10 × 0.5 = 5m.

μsmgcosθ = μs × m × 10 × cos30° = 0.6 × m × 10 × 0.866 = 5.196m.

Compare mgsinθ and μsmgcosθ:

Since mgsinθ (5m) < μsmgcosθ (5.196m), the static friction is sufficient to prevent motion.

Determine acceleration:

The block does not move, so its acceleration is 0 m/s².

∴ The acceleration of the block is zero.

Top Common forces in mechanics MCQ Objective Questions

We slip while walking on a path having pond scum or green algae because: 

  1. The inertia of motion
  2. The friction is zero
  3. The friction between the feet and the path is increased.
  4. The friction between the feet and the path is reduced.

Answer (Detailed Solution Below)

Option 4 : The friction between the feet and the path is reduced.

Common forces in mechanics Question 6 Detailed Solution

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The correct answer is The friction between the feet and the path is reduced.

Important Points

  • Friction is a force between two surfaces that are sliding, or trying to slide, across each other.
    • For example, when you try to push or pull luggage along the floor, friction makes this difficult.
  • Friction always works in the opposite of the direction in which the object is moving or trying to move.
  • There are four types of friction:
    • Static Friction: Static friction acts on objects when they are resting on a surface.
      • For example, hiking in the woods, there is static friction between shoes and the trail each time put down the foot.
      • Without this static friction, feet would slip out and making it difficult to walk.
    • Sliding Friction: Sliding friction is friction that acts on objects when they are sliding over a surface.
      • Sliding friction is weaker than static friction.
    • Rolling Friction: Rolling friction is friction that acts on objects when they are rolling over a surface.
      • Rolling friction is much weaker than sliding friction or static friction.
      • For example, ground transportation use wheels, including bicycles, cars, 4-wheelers, roller skates, scooters, skateboards, Ball bearings.
    • Fluid Friction: Fluid friction is friction that acts on objects that are moving through a fluid.
      • A fluid is a substance that can flow and take the shape of its container. Fluids include liquids and gases.

A heavy iron block is placed on the ground, a person tries to push the block with the force of 200 N and the block does not move. This means

  1. Friction force = Applied force
  2. Friction force > Applied force
  3. Friction force ≥ Applied force
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : Friction force ≥ Applied force

Common forces in mechanics Question 7 Detailed Solution

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Concept:

Friction Force:

It is a kind of resisting force that acts between two contact bodies to avoid relative motion between them.

  • It is formed due to the close interaction of charged particles at the surface of the bodies due to electromagnetic forces.
  • Mathematically it is represented as Friction force = μ N.
  • Where N = normal force = m × g
  • Normal force: It is a force that is exerted by the surface perpendicular to the body. It is a component of the contact force.

Frictional forces are of two kinds:

  1. Static friction (fs): When two bodies do not slip on each other, the force of friction is static friction. It can change its value until its max. Fmax ≥ fs
  2. Kinetic Friction (fk): When two bodies slip over each other the force of friction is called kinetic friction.
  • Coefficient of Kinetic friction (μk) < Coefficient Static friction (μs).

F1 J.S 18.5.20 Pallavi D1

  • Here, Applied force = friction force
  • Applied force (F) = μ N = μ m g

Explanation:

  • Since the block does not move so frictional force (fs) ≥ Applied force
  • This means frictional force (fs) ≥ 200 N

A person is standing in an elevator. In which situation he finds his weight less than actual when:

  1. The elevator moves upward with constant acceleration
  2. The elevator moves downward with constant acceleration
  3. The elevator moves upward with uniform velocity
  4. The elevator moves downward with uniform velocity

Answer (Detailed Solution Below)

Option 2 : The elevator moves downward with constant acceleration

Common forces in mechanics Question 8 Detailed Solution

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CONCEPT:

Pseudo Force:

  • A Pseudo force (also called a fictitious force, inertial force, or d’Alembert force) is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference frame, such as a rotating reference frame.
  • Pseudo force comes in effect when the frame of reference has started acceleration compared to a non-accelerating frame.

EXPLANATION:

F1 Prabhu.Y 13-07-21 Savita D2

  • A person having mass m is standing on a weighing machine which is placed in a stationary lift. In this case, the actual weight of a person is mg.
  • This weight on that weighing machine offers a normal reaction N, which is nothing but the reading of the weighing machine.
  • This reaction exerted by the weighing machine on the man is the apparent weight of the man.
  • This apparent weight depends upon the motion of the lift.

Case 1 :

  • When the lift is at rest or moving with a uniform velocity no matter upwards or downwards, the apparent weight of the man is given as,

​⇒ N = mg i.e., the actual weight of a man.

Case 2 :

  • When the lift accelerates upwards with an acceleration a, the apparent weight is given as,
  • ​From FBD we can see that,

⇒ N - mg = ma

⇒ N = m(g + a)

  • In this case, the apparent weight of the man has become more than the actual weight.

​Case 3 :

  • When the lift accelerates downward with an acceleration a, the apparent weight is given as,
  • ​From FBD we can see that,

⇒ mg - N = ma

⇒ N = m(g - a)

  • In this case, the apparent weight of the man has become less than the actual weight. Hence, option 2 is correct.

Which option is correct about the wheel of a car moving on the road?

  1. Rolling friction > Static friction > Kinetic friction
  2. Static friction > Kinetic friction > Rolling friction
  3. Static Friction > Rolling Friction > Kinetic Friction
  4. Rolling friction > Kinetic friction > Static friction

Answer (Detailed Solution Below)

Option 2 : Static friction > Kinetic friction > Rolling friction

Common forces in mechanics Question 9 Detailed Solution

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The correct option is Static friction > Kinetic friction > Rolling friction.

Key Points

  • Friction: The motion of any object is resisted by an opposing force known as the force of friction.
  • The rolling friction is very low as compared to the other two frictions.
  • If we compare kinetic and static friction then the value of static friction is more because it is present when the motion is not started after overcoming this friction object is able to move.
  • Static friction > Kinetic friction > Rolling friction

Additional Information

Friction is of three types:

  • Static Friction: The friction force encountered when the object is in the rest position that's static friction.
  • Kinetic Friction: If the object is sliding against each other then the friction present at that time is kinetic friction.
  • Rolling Friction: If the object is rolling on the surface the friction which is present at that time between the surfaces is known as Rolling friction.

A body of mass 20 kg is resting on a rough horizontal plane of the co-effcient of friction 0.5. If 40 N force is applied on a body then find the velocity of the body after 10 sec. (Take g = 10 m/sec2)

  1. 30 m/sec
  2. 20 m/sec
  3. 0 m/sec
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 0 m/sec

Common forces in mechanics Question 10 Detailed Solution

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CONCEPT:

Friction:

  • The resistance offered by the surfaces that are in contact with each other when they move over each other is called friction.
  • Factors affecting friction:
    1. Types of surfaces in contact.
    2. The normal force between the two surfaces.
  • Friction force does not depend on the velocity of the object.


Limiting friction:

  • The maximum friction that can be generated between two static surfaces in contact with each other.
  • Once a force applied to the two surfaces exceeds the limiting friction, the motion will occur.
  • We know that the limiting friction force between any two surfaces is given as,


⇒ F = μN

Where F = friction force, μ = coefficient of friction and N = normal reaction

Newton's second law of motion:

  • According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the applied unbalanced force.
  • The magnitude of the force is given as,


⇒ F = ma

Where m = mass and a = acceleration

CALCULATION:

Given:

 m = 20 kg, μ = 0.5, P = 40 N, t = 10 sec and g = 10 m/sec2

​Since the body is resting on a rough horizontal plane so the normal reaction is given,

⇒ N = mg

⇒ N = 20 × 10

⇒ N = 200 N

The limiting friction is given as,

⇒ F = μN

⇒ F = 0.5 × 200

⇒ F = 100 N

  • Since the applied force is less than the limiting friction so the body will remain at rest. Hence, option 3 is correct.

Which of the following forces is a contact force ?

  1. Force of gravity
  2. Magnetic force
  3. Force of friction
  4. Electrostatic force

Answer (Detailed Solution Below)

Option 3 : Force of friction

Common forces in mechanics Question 11 Detailed Solution

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CONCEPT:

  • Sir Isaac Newton gave us the concept of force.
  • Force is a stimulus provided to an object in order to make it do something.
  • Force can be both against the motion and for it.
  • The amount of force required is related to the mass of the object, the greater the mass, the greater the force required to move it. 

 

There are two types of forces based on their applications:

  1. Contact Force - It is a force applied to a body by another body that is in contact with it.
  • eg: Muscular Force, Mechanical Force, Frictional Force
  1. Non-Contact Force - It is a force that acts on an object without coming physically in contact with it.
  • eg: Gravitational Force, Electrostatic force, Magnetic force

EXPLANATION:

  • Force of gravity (GRAVITATIONAL FORCE):  This force is a force that attracts any two objects with mass.
    • Gravitational force is attractive because it always tries to pull masses together, it never pushes them apart. In fact, every object, including you, is pulling on every other object in the entire universe. This is called Newton's Law of Gravitation.
  • Magnetic force:
    • Attraction or repulsion that arises between electrically charged particles because of their motion.
    • It is the basic force responsible for such effects as the action of electric motors and the attraction of magnets for iron.
  • Electrostatic force: It is also known as the Coulomb force or Coulomb interaction. It's an attractive or repulsive force between two electrically charged objects. Like charges repel each other while unlike charges attract each other. Coulomb's law is used to calculate the strength of the force between two charges.

If a body is moving in such a way that force acting on it is uniform and perpendicular to the motion of the body, then: 

  1. Only speed will change
  2. Speed and direction of the body will change
  3. The velocity of the body does not change
  4. Speed remains the same, but direction changes 

Answer (Detailed Solution Below)

Option 4 : Speed remains the same, but direction changes 

Common forces in mechanics Question 12 Detailed Solution

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CONCEPT:

  • Circular motion: When a body moves in such a way that its force is always perpendicular to the velocity, the motion will be circular motion.
  • Centripetal force: The force that is perpendicular to the velocity and directed towards the center is known as centripetal force. or
    • The force that makes a body move in a circular motion is known as centripetal force.
    • The direction of this force is always perpendicular to the direction of the velocity.

The centripetal force acting on a body of mass 'm' revolving with radius 'r' is:

\(F = {mv^2\over r}\)

EXPLANATION:

Given that applied force is uniform and perpendicular to the motion.

  • From the definition of the centripetal force, if a force is perpendicular to the motion, it is a centripetal force and this force makes the body move in a circular motion.

We know \(F = {mv^2\over r}\)

  • Since F is uniform, so v will also be uniform. So speed (magnitude of velocity) will not change.
  • In a circular motion, a body moves in a circle, and in a circle the direction of motion changes at every point.
  • At point A, the direction of the velocity is in the upward direction. 
  • At point B, the direction of the velocity is in the leftward direction,
  • At point C, the direction of the velocity is in the downward direction.

  • The body comes in the initial direction after one complete circle.
  • So If the force on a moving body is always uniform and perpendicular to its motion, then Speed remains the same, but direction changes.
  • Hence the correct answer is option 4.

Which of the following is contact force/forces?

I. Friction force

II. Magnetic force

III. Gravitation force

  1. Only I
  2. Only II
  3. Only III
  4. Both II and III

Answer (Detailed Solution Below)

Option 1 : Only I

Common forces in mechanics Question 13 Detailed Solution

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The correct answer is Only I

  • The friction force is a contact force.
  • Contact force is a force that is applied by objects in contact with each other.
  • The contact force is governed by Newton’s Laws.
  • Friction is the resistance to motion of one object moving relative to another.
  • Magnetic force is the non-contact force.
  • Gravitation force is also a non-contact force.

Types of contact forces:

  • Frictional Force
  • Applied Force
  • Normal Force

If the two rough surfaces are polished and put in contact, the friction force between them will __________.

  1. increase
  2. decrease
  3. remain constant
  4. Either increase or decrease

Answer (Detailed Solution Below)

Option 2 : decrease

Common forces in mechanics Question 14 Detailed Solution

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CONCEPT:

  • Friction is the resistance force that opposes relative motion between two objects. 
    • It is not a fundamental force, like gravity or electromagnetism.
    • The direction of friction is always in such a way that it opposes the relative motion
  • The friction force between two surfaces is due to the roughness of the contact surface.

EXPLANATION:

  • The friction force between two surfaces is due to the roughness of the contact surface.
  • When the surfaces are polished, their roughness decreases.
  • They become smooth.
  • Hence the friction between them decreases.
  • So the correct answer is option 2.

If a man pulls a square-shaped box of mass 10 kg with the force of 50 Newton and body start to accelerate in the direction of force with 2 m/s2. What will be the value of the coefficient of friction and the type of frictional force?

  1. 0.3, Static friction
  2. 0.3, Kinetic friction
  3. 0.2, Static friction
  4. 0.2, Kinetic friction

Answer (Detailed Solution Below)

Option 2 : 0.3, Kinetic friction

Common forces in mechanics Question 15 Detailed Solution

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CONCEPT:

Friction Force: It is a kind of resisting force that acts between two contact bodies to avoid relative motion between them.

  • It is formed due to the close interaction of charged particles at the surface of the bodies due to electromagnetic forces.
  • Mathematically it is represented as Friction force = μ N.
  • Where N = normal force = m × g
  • Normal force: It is a force that is exerted by the surface perpendicular to the body. It is a component of the contact force.

Frictional forces are of two kinds:

  1. Static friction (fs): When two bodies do not slip on each other, the force of friction is static friction. It can change its value until its max. Fmax ≥ fs
  2. Kinetic Friction (fk): When two bodies slip over each other the force of friction is called kinetic friction.

F1 J.S 18.5.20 Pallavi D1

  • Coefficient of Kinetic friction (μk) < Coefficient Static friction (μs).
  • Force: It is a kind of push or pull of a body.
    • Force produces acceleration in a body.
    • F = m × a [m = mass of the body, a = acceleration in a body]
  • Free Body Diagram: It is an illustration to visualize the forces and resultants of each body respectively

CALCULATION:

  • Given, Mass of the body (m) = 10 kg, Normal force = mg = 10 × 10 = 100 N
  • Coefficient of friction (μ) = ?
  • Free body diagram of the box is

F1 J.S 18.5.20 Pallavi D3

Since the body accelerates in the direction of the force.

The laws of motion in the horizontal line is

F - μk N = ma

⇒ 50 – (μk × 100) = 10 kg × 2 m/s2

⇒ μ k = 0.3

Since the body is already in motion the kinetic friction is working.

So, kinetic friction with a value of 0.3 is working here.
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