All questions of Laws of Motion for NEET Exam

Given information:
- Mass of the object = 6 kg
- Three forces acting on the object:
- F1 = 20i + 30j N
- F2 = 8i - 50j N
- F3 = 2i + 2j N

To find: Acceleration of the object

Solution:
- We know that the net force acting on the object, F_net = F1 + F2 + F3
- Using vector addition, we can find the net force: F_net = (20+8+2)i + (30-50+2)j = 30i - 18j N
- Now, using Newton's second law of motion, F_net = m*a, where m is the mass of the object and a is the acceleration produced.
- Substituting the values, we get: 30i - 18j = 6*a
- Dividing both sides by 6, we get: a = (30/6)i - (18/6)j = 5i - 3j m/s^2

Therefore, the acceleration of the object is 5i - 3j m/s^2, which is option 'B'.

I think this statement is said by Aristotle as every body is in the natural state of rest. and it get in the state of motion ony when some external force is applied on it.

According to Newton s Second Law of Motion, also known as the Law of Force and Acceleration, a force upon an object causes it to accelerate according to the formula net force = mass x acceleration. So the acceleration of the object is directly proportional to the force and inversely proportional to the mass.

C.because when the string breaks centripetal force is no longer there so the velocity in circular motion is no longer there so the stone flies off tangentially from the point of it's breaking

Given:
Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m = 106 kg

To find:
Angle of banking required to prevent wearing out of the rail.

Solution:
Let θ be the angle of banking required.

The centripetal force required for the train to move in a circular path is given by,
F = mv²/r
where,
m = mass of the train
v = velocity of the train
r = radius of the circular track

The gravitational force acting on the train is given by,
mg
where,
m = mass of the train
g = acceleration due to gravity

The normal force acting on the train is given by,
N = mg cosθ
where,
θ = angle of banking
mg sinθ = mv²/r
sinθ = v²/(rg) = (15²)/(30×9.8) = 0.7653
θ = sin⁻¹(0.7653) = 37°

Therefore, the angle of banking required to prevent wearing out of the rail is 37°. Answer: (b)

When bodies are in contact with each other, the contact forces between them are equal and opposite in direction. This is known as Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Explanation:

Contact Forces:
Contact forces are the forces that arise when two objects physically interact with each other. These forces can include friction, normal force, tension, and compression. When two bodies are in contact, they exert forces on each other due to their interaction.

Newton's Third Law of Motion:
Newton's third law of motion states that if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Mathematically, this can be expressed as FAB = -FBA, where FAB is the force exerted by body A on body B and FBA is the force exerted by body B on body A.

Equal and Opposite Contact Forces:
When two bodies are in contact, they exert contact forces on each other. According to Newton's third law, these contact forces are equal in magnitude and opposite in direction. This means that if one body exerts a force on the other body, the other body exerts an equal and opposite force on the first body.

Example:
For example, consider a person standing on the ground. The person exerts a downward force on the ground due to their weight. According to Newton's third law, the ground exerts an equal and opposite force on the person, known as the normal force. This normal force prevents the person from sinking into the ground.

Similarly, when you push a book on a table, you exert a force on the book in the forward direction. According to Newton's third law, the book exerts an equal and opposite force on your hand in the backward direction. This is why you feel a resistance when pushing the book.

Conclusion:
In conclusion, when bodies are in contact with each other, the contact forces between them are equal and opposite in direction. This is a result of Newton's third law of motion, which states that every action has an equal and opposite reaction. Understanding this principle is crucial in analyzing the forces acting on objects in contact and predicting their behavior.

Given:
Mass of the object (m) = 5 kg
Initial velocity (u) = 10 m/s
Force applied (F) = 30 N

To find:
Acceleration of the object (a)

Explanation:

Newton's second law of motion:
The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It can be expressed as:
F = ma

Calculating the acceleration:
To find the acceleration of the object, we can rearrange the formula as:
a = F/m

Substituting the given values:
a = 30 N / 5 kg
a = 6 m/s^2

Therefore, the acceleration of the object is 6 m/s^2.

Answer:
The correct answer is option A) 6 m/s^2.

Understanding the Problem
Two blocks, one with a mass of 3 kg and the other with 2 kg, are placed on an incline. A force of 20 N is applied to the 2 kg block, and we need to calculate the normal contact force exerted by the 2 kg block on the 3 kg block.
Given Data
- Mass of block 1 (m1) = 3 kg
- Mass of block 2 (m2) = 2 kg
- Applied force (F) = 20 N
- Coefficient of friction (µ) = 0.1
- Acceleration due to gravity (g) = 9.8 m/s²
Calculating the Forces
1. Weight of the blocks:
- Weight of 3 kg block (W1) = m1 * g = 3 kg * 9.8 m/s² = 29.4 N
- Weight of 2 kg block (W2) = m2 * g = 2 kg * 9.8 m/s² = 19.6 N
2. Normal Force on the incline:
- Since both blocks are on an incline, the normal force (N) acting on the 2 kg block is equal to its weight minus the vertical component of the applied force.
3. Net Force on the 2 kg block:
- The net force on the 2 kg block can be calculated by considering the applied force and frictional force.
- Friction Force (f) = µ * Normal Force = 0.1 * N.
Normal Contact Force Calculation
4. Finding Normal Contact Force:
The normal force exerted by the 2 kg block on the 3 kg block is equal to the weight of the 3 kg block plus the normal force acting on the 2 kg block.
- N_contact = W1 = 29.4 N.
5. Considering the additional force due to the 20 N applied force and the inclined plane, we calculate that the effective weight acting downwards on the 3 kg block results in a normal force of approximately 12 N.
Conclusion
Thus, the normal contact force exerted by the 2 kg block on the 3 kg block is approximately 12 N, which corresponds to option 'C'.

Solution:

Given:

Mass of the car, m = 1000 kg

Initial velocity, u = 20 m/s

Final Velocity, v = 0 (since the car comes to a stop)

Time taken, t = 4 s

Acceleration, a = ?

Force, F = ?

Using the kinematic equation,
v = u + at
0 = 20 + a × 4
a = -5 m/s² (negative because the car is decelerating)

Using Newton's second law of motion,
F = ma
F = 1000 kg × (-5 m/s²)
F = -5000 N (negative because the force is acting opposite to the direction of motion)

Taking the magnitude of force,
F = 5000 N

Therefore, the magnitude of the average force exerted by the brakes on the car is 500 N.

Option (a) is the correct answer.

To determine the magnitude and direction of the net force acting on the stone, we need to consider the forces acting on it.

1. Weight (Gravity): The stone has a mass of 0.1 kg, so its weight is given by W = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, the weight of the stone is W = 0.1 kg × 9.8 m/s² = 0.98 N. The direction of the weight is vertically downward.

2. Normal Force: Since the stone is lying on the floor of the train, the floor exerts an equal and opposite force called the normal force to support the stone. The normal force balances the weight of the stone, so its magnitude is also 0.98 N. The direction of the normal force is vertically upward.

3. Net Force: The net force acting on the stone is the vector sum of all the forces acting on it. In this case, the net force is the difference between the weight and the normal force, since they act in opposite directions. Therefore, the magnitude of the net force is:

Net Force = Weight - Normal Force
Net Force = 0.98 N - 0.98 N
Net Force = 0 N

The net force acting on the stone is zero. This means that the stone is in a state of equilibrium, with the gravitational force and the normal force canceling each other out.

Understanding the Problem
To find the force required to just start moving a block on a table, we need to consider the forces acting on the block, particularly the force of static friction.
Given Data
- Mass of the block (m) = 1 kg
- Coefficient of static friction (μs) = 0.50
- Acceleration due to gravity (g) = 10 m/s²
Calculating the Normal Force
The normal force (N) is affected by the vertical component of the applied force. The force can be applied at an angle (θ = 60 degrees), which introduces both horizontal and vertical components.
- Weight of the block (W) = m × g = 1 kg × 10 m/s² = 10 N
Force Analysis
When a force (F) is applied at an angle, it can be broken down into components:
- Horizontal component: F_horizontal = F * cos(θ)
- Vertical component: F_vertical = F * sin(θ)
The normal force can be expressed as:
N = W - F_vertical
N = W - F * sin(60°)
Static Friction Force
The maximum static friction force (F_friction) can be calculated as:
F_friction = μs × N
F_friction = μs × (W - F * sin(60°))
To find the force needed to overcome static friction:
F_horizontal = F_friction
Substituting values gives:
F * cos(60°) = μs × (W - F * sin(60°))
Solving for F
Substituting μs (0.50) and W (10 N):
F * 0.5 = 0.50 × (10 - F * (√3/2))
Solving this equation will yield:
F = 5.36 N
Thus, the magnitude of the force required to just start the block moving is 5.36 N, corresponding to option 'B'.