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A passenger is travelling in a.train which is moving at 40 \(\mathrm{m} / \mathrm{s}\). His suitcase is kept on the berth. The driver of the train applies breaks such that the speed of the train decreases at a constant rate to \(20 \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\). What should be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not to slide during retardation of the train? a. \(0.3\) b. \(0.5\) c. \(0.1\) d. \(0.2\)

Short Answer

Expert verified
The minimum coefficient of friction is 0.5.

Step by step solution

01

Understand the Context

The train is slowing down, which means it is undergoing negative acceleration (deceleration). The suitcase on the berth experiences this deceleration as well.
02

Determine the Deceleration

To find the deceleration, use the formula for acceleration: \( a = \frac{{v_f - v_i}}{t} \). Here, \( v_i = 40 \ \mathrm{m/s} \), \( v_f = 20 \ \mathrm{m/s} \), and \( t = 5 \ \mathrm{s} \).Substitute these values to find \( a \):\[ a = \frac{{20 - 40}}{5} = \frac{{-20}}{5} = -4 \ \mathrm{m/s^2} \]The negative sign indicates deceleration.
03

Apply Newton's Second Law

To prevent the suitcase from sliding, the frictional force must be equal to or greater than the force due to deceleration. Using Newton's second law, force exerted by friction is \( F_f = ma \), where \( m \) is the mass of the suitcase and \( a = -4 \ \mathrm{m/s^2} \).
04

Frictional Force Requirement

The frictional force is given by \( F_f = \mu mg \), where \( \mu \) is the coefficient of friction and \( g \,=\, 9.8 \, \mathrm{m/s^2} \) is the acceleration due to gravity.
05

Equate Forces and Solve for \( \mu \)

To prevent sliding, set the frictional force equal to the deceleration force and solve for \( \mu \):\[ \mu mg = ma \]\[ \mu g = a \Rightarrow \mu = \frac{|a|}{g} \]Substitute \( a = 4 \ \mathrm{m/s^2} \) and \( g = 9.8 \ \mathrm{m/s^2} \):\[ \mu = \frac{4}{9.8} \approx 0.41 \]
06

Determine the Correct Option

Compare \( \mu = 0.41 \) with the provided options. The closest option higher than 0.41 ensures the suitcase does not slide. The answer is therefore \( b. 0.5 \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that describes how the motion of an object is affected by forces. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
This can be expressed with the formula:
  • \( F = ma \)
In this context, a suitcase on a berth in a train experiences a force due to the train's deceleration. As the train slows down, the negative acceleration, or deceleration, occurs. The negative sign indicates the direction of acceleration is opposite to the initial movement of the train.
Newton's Second Law helps in calculating the forces needed to keep the suitcase stationary relative to the train. The law indicates that if a sufficient opposing force is not applied (like friction), the suitcase will start to move.
Deceleration
Deceleration is simply the negative acceleration. It happens when an object reduces its speed. There can be different causes for deceleration, such as applying brakes to a moving object.
To calculate deceleration, the formula used is:
  • \( a = \frac{v_f - v_i}{t} \)
where:
  • \( v_i \) is the initial velocity
  • \( v_f \) is the final velocity
  • \( t \) is the time taken for the change in velocity
In the train situation, the initial speed is 40 m/s, and the final speed is 20 m/s over a period of 5 seconds. By substituting these into our formula, we find that the deceleration (or the rate of slowing down) is \(-4 \, \text{m/s}^2\).
This value helps in understanding how quickly the object is stopping and ensures we can calculate the necessary conditions to maintain equilibrium, like through friction.
Coefficient of Friction
The coefficient of friction, often denoted by \( \mu \), is a measure of how much frictional force exists between two surfaces. It plays a crucial role in determining whether an object maintains its position when forces are applied.
This dimensionless number ranges between 0 and 1 (though it can be higher for very sticky surfaces). In physics problems, the coefficient of friction helps decide whether an object will slide over a surface or remain stationary.
For the train exercise, the coefficient of friction tells us whether the suitcase will stay in place during the train's deceleration. We derive it using the equation:
  • \( \mu = \frac{|a|}{g} \)
With \( |a| = 4 \, \text{m/s}^2 \) and \( g = 9.8 \, \text{m/s}^2 \), the calculated \( \mu \) is approximately 0.41. Thus, comparing it to the choices provided, only a coefficient higher than 0.41 will ensure the suitcase does not slide, making 0.5 the safest choice.

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Most popular questions from this chapter

A man is raising himself and the crate on which he stands with an acceleration of \(5 \mathrm{~m} / \mathrm{s}^{2}\) by a massless rope-andpulley arrangement. Mass of the man is \(100 \mathrm{~kg}\) and that of the crate is \(50 \mathrm{~kg}\). If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), then the tension in the rope is a. \(2250 \mathrm{~N}\) b. \(1125 \mathrm{~N}\) c. \(750 \mathrm{~N}\) d. \(375 \mathrm{~N}\)

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