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Chapter 5: Problem 18

A block of mass \(50 \mathrm{~kg}\) slides over a horizontal distance of \(1 \mathrm{~m}\). If the coefficient of friction between their surfaces is o.2, then work done against friction is [CBSE PMT 1999, 2000; AIIMS 2000; BHU 2001] (a) \(98 J\) (b) \(72 . J\) (c) \(56 J\) (d) \(34 J\)

Short Answer

Expert verified
The work done against friction when the block is moved for 1 m is \(98 J\). Hence, the correct answer is (a) \(98 J\).

Step by step solution

01

Determine Force of Friction

Frictional force can be calculated using the formula \(F = \mu * m * g\), where \(\mu\) is the coefficient of friction, \(m\) is the mass of the block, and \(g\) is the acceleration due to gravity. Plugging in the given values, \(F = 0.2 * 50 * 9.8 = 98 N\).
02

Calculate Work Done

The work done against friction is calculated by multiplying the force of friction by the distance over which the block is moved. In this case, the formula is \(W = F * d\). So, the work done is \(W = 98 N * 1 m = 98 J\) (Joules).
03

Check Against Given Options

The calculated work done is 98 Joules. Comparing this result with the options given in the problem, it matches with option (a), hence, (a) \(98 J\) is the correct answer.

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Key Concepts

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

Friction
Friction is a force that resists the sliding or rolling of one object over another. It is crucial in everyday life and mechanical systems. For instance, walking or driving involves friction between your feet or tires and the ground that allows motion.
Friction can be categorized as:
  • Static friction: Resists the initiation of motion between two surfaces.
  • Kinetic friction: Acts on objects already in motion.
The magnitude of frictional force is proportional to the normal force and is influenced by how rough or smooth the surfaces are. The formula is: \[ F_{friction} = \mu imes N \]where,
\( \mu \) = coefficient of friction (no unit)
\( N \) = normal force (equal to weight for horizontal surfaces).
It is vital to note that friction always acts opposite to the direction of motion or intended motion. In the problem, friction is responsible for opposing the movement of the block along the surface.
Newton's Laws of Motion
Newton's Laws of Motion form the foundation of classical mechanics and dictate how objects move in response to forces.
  • First Law: An object will remain at rest, or in uniform motion in a straight line, unless acted upon by an external force. This is often referred to as the law of inertia.
  • Second Law: The force on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)). This law helps in quantifying the effect of forces.
  • Third Law: For every action, there is an equal and opposite reaction. This means that forces always occur in pairs.
In the exercise, applying Newton's Second Law helps in understanding how the frictional force, calculated using the coefficient of friction and the block's mass, opposes the motion and hence does work on it.
Kinetic Friction
When two surfaces slide against each other, kinetic friction comes into play. This type of friction acts while the object is in motion and typically lower than static friction.
Kinetic friction is defined by:
  • Equation: \[ F_{kinetic} = \mu_k \times N \]where,
    \( \mu_k \) = coefficient of kinetic friction
  • Characteristics:
    • Less than the maximum static friction force.
    • Depends on the nature of the surfaces and the force pressing them together.
Kinetic friction explains why once an object starts moving, it continues with less applied force. In our exercise, the force of kinetic friction is computed to find out how much work is needed to overcome this resisting force over a distance, leading to a calculation of 98 Joules.

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Most popular questions from this chapter

If a ladder weighing \(250 \mathrm{~N}\) is placed against a smooth vertical wall having coefficient of friction between it and floor is \(0.3\), then what is the maximum force of friction available at the point of contact between the ladder and the floor (a) \(75 \mathrm{~N}\) (b) \(50 \mathrm{~N}\) (c) \(35 \mathrm{~N}\) (d) \(25 \mathrm{~N}\)

A body of mass \(2 k g\) is kept by pressing to a vertical wall by a force of \(100 \mathrm{~N}\). The friction between wall and body is o.3. Then the frictional force is equal to [Orissa JEE 20o3] (a) \(6 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(600 \mathrm{~N}\) (d) \(700 \mathrm{~N}\)

A block of mass \(10 \mathrm{~kg}\) is placed on a rough horizontal surface having coefficient of friction \(\mu=0.5\). If a horizontal force of \(100 N\) is acting on it, then acceleration of the block will be (a) \(0.5 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(5 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(15 \mathrm{~m} / \mathrm{s}^{2}\)

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 o.5, with what acceleration will the fireman slide down \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right) \quad\) [Pb. PMT 2002] (a) \(1 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(5 \mathrm{~m} / \mathrm{s}^{2}\)

A block of mass \(1 \mathrm{~kg}\) slides down on a rough inclined plane of inclination \(60^{\circ}\) starting from its top. If the coefficient of kinetic friction is \(0.5\) and length of the plane is \(1 m\), then work done against friction is (Take \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) ) \(\quad\) [AFMC 2000; KCET (Engg./Med.) 2001] (a) \(9.82 J\) (b) \(4.94 \mathrm{~J}\) (c) \(2.45 J\) (d) \(1.96 J\)
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