91Ó°ÊÓ

A train starts from rest at \(t=0\) along a straight track with a constant acceleration of \(5 \mathrm{~m} / \mathrm{s}^{2}\). A passenger at rest in train observes a particle of mass \(1 \mathrm{~kg}\) on the floor with which it has a co-efficient of friction \(\mu_{s}=\mu_{k}=0.6 .\) At \(t=4 s\), a horizontal force \(F=13 \mathrm{~N}\) is applied on the particle for \(2 \mathrm{~s}\) duration. The passenger observes that the particle is now moving in a perpendicular direction of motion of the train. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) The kinetic energy of the particle at \(t=20 \mathrm{~s}\) with respect to the ground is (A) \(5 \times 10^{3} \mathrm{~J}\) (B) \(6 \times 10^{3} \mathrm{~J}\) (C) \(8 \times 10^{3} \mathrm{~J}\) (D) \(7 \times 10^{3} \mathrm{~J}\)

Step by step solution

01

Determine the acceleration and final velocity of the train

Since the train starts from rest and accelerates at a constant rate of 5 m/s², we can find the final velocity of the train at time t = 20 s using the following formula: Final velocity (v) = Initial velocity (u) + Acceleration (a) × Time (t) Since the train starts from rest, its initial velocity is 0. Therefore, v = 0 + (5 m/s²) × (20 s) = 100 m/s

02

Find the maximum static friction

The maximum static friction can be calculated using the formula: f_max = μs × m × g, where μs is the coefficient of static friction, m is the mass of the particle, and g is the acceleration due to gravity. From the given information, f_max = (0.6) × (1 kg) × (10 m/s²) = 6 N

03

Find the acceleration of the particle with respect to the train

Since the force F = 13 N applied on the particle is greater than the maximum static friction (6 N), the particle starts moving. The net force acting on the particle is (F - f_max) = (13 N - 6 N) = 7 N Using Newton's second law, we can find the acceleration of the particle with respect to the train: Net force = m × a a_particle_train = Net force/ mass = (7 N) / (1 kg) = 7 m/s²

04

Find the velocity of the particle with respect to the train at t = 20 s

When the force is applied for 2 s, the particle's final velocity with respect to the train can be calculated using the following formula: v_particle_train = u_particle_train + (a_particle_train × t_force) Since the particle is at rest initially, v_particle_train = 0 + (7 m/s² × 2 s) = 14 m/s

05

Find the particle's velocity with respect to the ground

Since the particle is moving perpendicular to the train's motion, we can calculate its velocity with respect to the ground using the Pythagorean theorem: v_particle_ground = √(v_train² + v_particle_train²) Substituting the values we calculated earlier, v_particle_ground = √((100 m/s)² + (14 m/s)²) = √(10000 + 196) = √10196 = 101 m/s

06

Calculate the kinetic energy

Finally, we can find the kinetic energy of the particle with respect to the ground using the formula: Kinetic energy = 1/2 × m × v², KE = 1/2 × (1 kg) × (101 m/s)², KE = 0.5 × 101 × 101 = 0.5 × 10201 = 5100.5 J Rounding to the closest whole number, the kinetic energy of the particle is approximately 5101 J, which is closest to 5 × 10³ J in the given options. So the answer is (A) 5 × 10³ J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

acceleration

Acceleration is a fundamental concept in physics that describes how quickly an object speeds up or slows down. When an object experiences constant acceleration, it means that it changes its velocity by the same amount each second. In this exercise, the train starts with an acceleration of 5 meters per second squared (m/s²). This simply means that for every second, the train's speed increases by 5 m/s. Because the train is initially at rest, its initial velocity is 0 m/s. After 20 seconds, as calculated, the train's velocity reaches 100 m/s due to its constant acceleration.

• Acceleration helps determine the final speed of an object.
• It is measured in units of \( m/s^2 \).
• An object moving with constant acceleration follows a linear velocity-time graph.

Understanding acceleration is crucial for calculating different parameters such as the final velocity and displacement, which are often needed in dynamics problems.

static friction

Static friction is a type of friction that keeps an object at rest when it is subjected to a force. It must be overcome before an object can start moving. In the problem, the coefficient of static friction between the particle and the train's floor is given as 0.6. This coefficient is a measure of how much resistance exists before motion begins.

• Static friction prevents motion up to a maximum value.
• The maximum static friction force can be calculated using the formula \( f_\mathrm{max} = \mu_s \times m \times g \), where \( \mu_s \) is the coefficient of static friction, \( m \) is mass, and \( g \) is gravitational acceleration.
• In this scenario, it is 6 N, as calculated using \( \mu_s = 0.6, m = 1 \mathrm{kg}, g = 10 \mathrm{m/s}^2 \).

When the force exceeds this static friction, the object starts to move, transitioning to kinetic friction. Static friction is an essential concept for understanding how forces interact with stationary objects.

Newton's second law

Newton's second law of motion is a key principle in physics that explains how the velocity of an object changes when it is subjected to a net force. This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It is mathematically expressed as \( F = m \times a \).

• According to the law, if multiple forces act on an object, the net force can be determined by vector addition of all the forces.
• Once the net force is known, acceleration can be calculated.
• In our problem, after accounting for static friction, the particle experienced a net force of 7 N, leading to an acceleration of 7 m/s² (since the mass of the particle is 1 kg).

This law can be used to determine how objects accelerate, how they change direction, and how forces interact in dynamic systems. It's fundamental for solving many physics problems dealing with motion.