Force and Acceleration Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Define Instantaneous acceleration.

Ans. The acceleration of a particle at a particular instant of time during its motion is called instantaneous acceleration.

It is given as the limit of average acceleration at the time interval Δt goes to zero.

a = lim _Δt→0 \(\frac{ \Delta v}{\Delta t} = \frac{dv}{dt}\)

Ques. What is the unit of force in the CGS system of unit?

1. Newton
2. erg
3. Coulomb
4. dyne

Ans. The correct option is d. Dyne

Explanation: The unit of force in the CGS system is Dyne.

1 dyne = 1 gram x 1cm s^-2

1 Newton = 10⁵ dyne

Ques. Write the unit and dimensional formula of acceleration.

Ans. Acceleration is defined as the rate of change of velocity.

SI unit of acceleration is meter-per-second square (m/s²).

The dimensional formula of acceleration is [ M⁰ L T^-2 ]

Ques. How much net force is required to accelerate a body of mass 60 kg with an acceleration of 6 m/s² ?

1. 540 N
2. 360 N
3. 130 N
4. 260 N

Ans. The correct option is b. 360 N

Explanation: According to Newton’s second law of motion

F = ma

F = 60 x 6 = 360 kg m/s² = 360 N

Ques. Define the force of friction.

Ans. Friction is the force that opposes the relative motion or the tendency of relative motion between two bodies in contact and acts tangentially along the surface of contact.

Ques. Define the law of inertia.

Ans. According to this law, Inertia is an inherent property of the body that has the inability to change its state of rest or uniform motion or change the direction of motion by itself.

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Short Answers Questions [2 Marks Questions]

Ques. Define force. Write the effect of force.

Ans. Force is a physical quantity that causes motion, opposes the motion, or changes the direction of motion of a body.

The effect of force is

• It may change the state of rest or of uniform motion of the body.
• It may change the direction of the body.
• It may change the shape and size of the body.

Ques. A bullet of 200 g was fired from a gun moving with a velocity of 20 m/s and hit a wooden log. The bullet stops after traveling a distance of 40 cm in the wooden log. Calculate the retarding force exerted by the log on the bullet.

Ans. Given

• Mass of the bullet, m = 200 g = 0.2 kg
• Initial velocity of the bullet, u = 20 m/s
• Final velocity, v = 0
• Distance covered, S = 40 cm = 0.4 m

Using the equation of motion, v² = u² + 2aS, we get

Acceleration of the bullet, a = (v² - u²)/2S = (0 - 20²)/(2 x 0.4) = - 500 m/s²

Retarding force on the bullet, F = ma = 0.2 x (-500) = - 100 N

Ques. Define Centriprtal acceleration.

Ans. When a particle is moving along a circular path, its direction changes continuously. Due to the change in the direction of motion, its velocity changes, and hence the particle is accelerated. This acceleration is called centripetal acceleration.

Ques. Define Centrifugal force.

Ans. A force that tends to move the bodies or particles of a rotating frame, away from the axis of rotation is called centrifugal force.

The magnitude of centrifugal force is equal to the magnitude of centripetal force.

Read More: Angular Acceleration Formula

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Long Answers Questions [3 Marks Questions]

Ques. What is the significance of Newton’s law of motion?

Ans. The significance of Newton’s law of motion is

• The first law of motion talks about the natural state of the motion of the body.
• The second law of motion says that if a body is not following its natural state of motion, then there has to be a net unbalanced external force acting on the body.
• The third law of motion talks about the nature of the force i.e. the force exists in pairs.

Ques. A mass of 5 kg is acted upon by a force of 1 N. Starting from rest, how much is the distance covered by the mass in 10 seconds?

Ans. Given

• Mass, m = 5 kg
• Force, F = 1 N
• Time, t = 10 s

Acceleration produced, a = F/m = 1/5 m/s²

Since the body is initially at rest, therefore initial velocity, u = 0

Using the equation of motion, S = ut + 1/2 at² we get

Distance covered, S = 0 + (1/2) x (1/5) x 10²

⇒ S = 10 m

Ques. Write Newton’s second law of motion and hence prove F = ma.

Ans. According to Newton’s second law of motion, the rate of change of momentum of a body is directly proportional to the applied force and it takes place in the direction in which the force acts.

Mathematically

F ∝ Δp/Δt

⇒ F = k Δp/Δt

Where k is the constant of proportionality.

Taking the limit Δt tends to zero, the term Δp/Δt becomes derivates of p with respect to t i.e. dp/dt. Therefore

F = k dp/dt

As, p = mv

⇒ F = k d(mv)/dt = km dv/dt

But, dv/dt = a (acceleration)

⇒ F = k ma

The value of constant of proportionality is considered as 1 for simplicity. Therefore we get

 F = ma

Ques. Define centripetal force. Give an example.

Ans. The force which deviates a moving particle from its linear path to move it along a circular path and acts along the radius and towards the center of the circle is called centripetal force.

The motion of planets around the sun is an example of centrifugal force. The gravitational force of the sun provides the necessary centripetal force to move the planets around the sun.

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Very Long Answers Questions [5 Marks Questions]

Ques. A small stone of mass 50 grams is thrown vertically upwards. Calculate the magnitude and direction of the net force on the stone

1. During its upward motion
2. During its downward motion
3. At its highest point

Is the magnitude and direction of the force on the stone change, if the stone were thrown at an angle of 450 with the horizontal?

Ans. Given

• Mass of the stone, m = 50 g = 50 x 10^-3 kg
• Acceleration = - acceleration due to gravity = -g = - 9.8 m/s²

1. During upward motion, the net force acting on the stone 

F = ma

⇒ F = (50 x 10^-3) x (- 9.8) = - 0.49 N

The direction of this force is vertically downward

2. During download motion, the net force acting on the stone 

F = ma

⇒ F = (50 x 10^-3) x (- 9.8) = - 0.49 N

The direction of this force is vertically downward

100. At the highest point, the net force acting on the stone 

F = ma

⇒ F = (50 x 10^-3) x (- 9.8) = - 0.49 N

The direction of this force is vertically downward.

If the stone were thrown at an angle of 450 with the horizontal, then at any point magnitude of the acceleration is equal to the acceleration due to gravity that acts vertically downward.

Hence the magnitude and direction of force remain the same.

Ques. A body of mass 30 kg stands on a weighing machine lying on the floor of a lift. What will be the weight of the body when

1. The lift moves upward with acceleration, a = 2.2 m/s²
2. The lift moves downward with acceleration, a = 2.2 m/s²
3. The lift falls freely

Ans. The weight of a body in a weighing machine is a measure of the normal reaction exerted by the weighing machine on the body. It is also known as the apparent weight of the body.

It may be zero or it may be equal, greater, or less than the true of the body.

Given 

• Mass of the body, m = 30 kg
• Acceleration, a = 2.2 m/s²

1. When the lift moves upward, then the apparent weight of the body is given by

W = mg + ma = m (g + a)

⇒ W = 30 (9.8 + 2.2) = 360 N

⇒ W = 360/9.8 = 36.7 kg

2. When the lift moves downward, then the apparent weight of the body is given by

W = mg - ma = m (g - a)

⇒ W = 30 (9.8 - 2.2) = 228 N

⇒ W = 228/9.8 = 23.3 kg

100. When the lift falls freely, then the apparent weight of the body is given by

W = m (g - a)

In free-fall condition, g = a 

⇒ W = m (g - g) = 0

The body will be in a state of weightlessness.

Ques. A block of mass 5 kg is moving with a speed of 8 m/s. A force of 10 N is applied to it for 3 seconds. Find the final speed of the block if force is applied

1. Along the motion
2. Opposite to the motion
3. Normal to the initial direction of the block

Ans. Given

• Mass of the block, m = 5 kg
• The initial speed of the block, u = 8 m/s
• Applied force, F = 10 N
• The time during which the force is applied, t = 3 s

Acceleration produced by the applied force on the block

a = F/m = 10/5 = 2 m/s²

The direction of acceleration is taken to be positive if the motion of the block is along the direction of motion and the direction of acceleration is taken to be negative if the motion of the block is opposite to the direction of motion.

1. When force is applied along the direction of motion, then final speed v is given by using the equation of motion

v = u + at

In this case, acceleration is taken as, a = +2 m/s²

On substituting the value of acceleration

⇒ v = 8 + (2 x 3) = 14 m/s

2. When force is applied in a direction opposite to the direction of motion, then final speed v is given by using the equation of motion

v = u + at

In this case, acceleration is taken as, a = -2 m/s²

On substituting the value of acceleration

⇒ v = 8 + (-2 x 3) 

⇒ v = 8 - 6 = 2 m/s

100. When the force acts perpendicular to the direction of motion.

Let the block is moving along X-axis and the acceleration be along Y-axis, then

Initial velocity, u = 8 m/s \(\hat{i}\)

Acceleration, a = 2 m/s²\(\hat{j}\)

Using the equation of motion

v = u + at

v = (8 m/s)\(\hat{i}\)+(2 m/s²)\(\hat{j}\)³

v = 8\(\hat{i}\)+6\(\hat{j}\)

Magnitude of final speed, v=8²+6²=10 m/s

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