/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A fireman of mass \(60 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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A fireman of mass \(60 \mathrm{~kg}\) slides down a pole He is pressing the pole with a force of \(600 \mathrm{~N}\). The coefficient of friction between the hands and the pole is \(0.5\) with what acceleration with the fireman slide down? \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(1 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(5 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The acceleration is \(5 \text{ m/s}^2\), so the answer is (d).

Step by step solution

01

Determine the Gravitational Force

First, we need to calculate the gravitational force acting on the fireman. This force can be found using the formula:\[ F_g = m imes g \]where \( m = 60 \text{ kg} \) is the mass of the fireman and \( g = 10 \text{ m/s}^2 \) is the acceleration due to gravity.\[ F_g = 60 \times 10 = 600 \text{ N} \]
02

Calculate the Frictional Force

The frictional force can be calculated using the formula:\[ F_f = \mu \times F_n \]where \( \mu = 0.5 \) is the coefficient of friction, and \( F_n = 600 \text{ N} \) is the normal force exerted by the fireman pressing the pole.\[ F_f = 0.5 \times 600 = 300 \text{ N} \]
03

Determine the Net Force Acting on the Fireman

The net force acting on the fireman can be found by subtracting the frictional force from the gravitational force:\[ F_{net} = F_g - F_f \]Inserting the values, we get:\[ F_{net} = 600 - 300 = 300 \text{ N} \]
04

Calculate the Acceleration

Using Newton's second law, we can find the acceleration by dividing the net force by the mass of the fireman:\[ a = \frac{F_{net}}{m} \]\[ a = \frac{300}{60} = 5 \text{ m/s}^2 \]
05

Choose the Correct Answer

Now, compare the calculated acceleration with the given options. The acceleration of the fireman sliding down is \(5 \text{ m/s}^2\).

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Key Concepts

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

Newton's Second Law
In physics, Newton's second law of motion is fundamental to understanding how forces affect motion. This law can be summarized by the equation: \[ F_{net} = m \cdot a \]where:
  • \( F_{net} \) is the net force acting on an object, measured in newtons (N).
  • \( m \) is the mass of the object in kilograms (kg).
  • \( a \) is the acceleration in meters per second squared \( \, \mathrm{m/s^2} \).
Newton's second law explains that the acceleration of an object depends directly on the net force acting on it and is inversely proportional to its mass. This means that for a constant mass, increasing the net force will increase the acceleration. Conversely, for a constant force, increasing the mass will decrease the acceleration.
In practical terms, when considering a fireman sliding down a pole, the net force acting on the fireman results from the difference between the gravitational force pulling him down and the frictional force acting against that motion. Understanding this relationship is crucial for calculating the fireman's acceleration, as seen in the exercise above.
Coefficient of Friction
The coefficient of friction, often represented by the Greek letter \( \mu \), plays a key role in understanding how surfaces interact with one another. This dimensionless number describes the ratio between the force of friction between two bodies and the force pressing them together, known as the normal force. The basic formula for frictional force \( F_f \) is:\[ F_f = \mu \cdot F_n \]where:
  • \( \mu \) is the coefficient of friction, which varies depending on the materials involved.
  • \( F_n \) is the normal force (in newtons).
In our fireman example, the coefficient of friction is 0.5. This means that the friction force is half of the normal force exerted by the fireman. The larger the coefficient, the greater the force needed to move one object over another. Coefficient of friction is crucial for solving problems involving movement and sliding, like when the fireman descends the pole. Understanding this value helps in calculating the force required to overcome the friction and determine the resulting acceleration.
Gravitational Force
Gravitational force is a universal force that draws objects toward one another, most notably pulling objects toward the center of the Earth. This force can be calculated using the formula:\[ F_g = m \cdot g \]where:
  • \( F_g \) is the gravitational force measured in newtons \( \, (N) \).
  • \( m \) is the mass of the object in kilograms (kg).
  • \( g \) is the acceleration due to gravity, which on Earth is approximately \( 10 \, \mathrm{m/s^2} \).
In the context of our exercise, the fireman's gravitational force is calculated by multiplying his mass (60 kg) by Earth's gravitational pull (10 m/s²), which gives a force of 600 N. This force is the downward pull experienced by the fireman as he slides down. Understanding gravitational force is fundamental in calculating the net force and subsequent acceleration in motion problems, illustrating how gravity acts as the primary force against which other forces must be balanced.

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Most popular questions from this chapter

A cylinder roll up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of frictional force acting on the cylinder are (a) up the inclined while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending

A \(1000 \mathrm{~kg}\) lift is supported by a cable that can support \(2000 \mathrm{~kg}\). The shortest distance in which the lift can be stopped when it is descending with a speed of \(2.5 \mathrm{~ms}^{-1}\) is [Take \(\left.g=10 \mathrm{~ms}^{-2}\right]\) (a) \(1 \mathrm{~m}\) (b) \(2 \mathrm{~m}\) (c) \(\frac{5}{32} \mathrm{~m}\) (d) \(\frac{5}{16} \mathrm{~m}\)

A body weighs \(8 \mathrm{~g}\) when placed in one pan and \(18 \mathrm{~g}\). when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, the true weight of the body is. (a) \(13 \mathrm{~g}\) (b) \(12 \mathrm{~g}\) (c) \(15.5 \mathrm{~g}\) (d) \(15 \mathrm{~g}\)

A smooth inclined plane of length \(L\) having inclination \(\theta\) with the horizontal is inside a lift which is moving down with retardation \(a\). The time taken by a body to slide down the inclined plane, from rest, will be (a) \(\sqrt{\frac{2 L}{(g+a) \sin \theta}}\) (b) \(\sqrt{\frac{2 L}{(g-a) \sin \theta}}\) (c) \(\sqrt{\frac{2 L}{g \sin \theta}}\) (d) \(\sqrt{\frac{2 L}{a \sin \theta}}\)

Two masses \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\) are connected at the two ends of a light inextensible string that goes over a frictionless pulley. The acceleration of the masaes and the tension in the string when the masses are released, are respectively [NCERT] (a) \(2 \mathrm{~m} / \mathrm{s}^{2}\) and \(90 \mathrm{~N}\) (b) \(4 \mathrm{~m} / \mathrm{s}^{2}\) and \(90 \mathrm{~N}\) (c) \(2 \mathrm{~m} / \mathrm{s}^{2}\) and \(60 \mathrm{~N}\) (d) \(4 \mathrm{~m} / \mathrm{s}^{2}\) and \(99 \mathrm{~N}\)

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