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Chapter 7: Problem 4

A particle of mass \(m\) slides a distance \(d\) down a plane inclined at \(\theta\) to the horizontal. The work done by the normal reaction \(R\) is: (a) \(R d\) (b) \(m g d \cos \theta\) (c) 0 (d) \(m g d \sin \theta\).

Short Answer

Expert verified
The work done by the normal reaction force is 0.

Step by step solution

01

Understand the Problem

Identify what is asked in the problem. We need to find the work done by the normal reaction force on a particle sliding down an inclined plane.
02

Identify the Normal Reaction Force

The normal reaction force, denoted as R, acts perpendicular to the inclined plane.
03

Work Done by Normal Force

Recall that work done by a force is given by the dot product of the force and the displacement vectors, which can also be expressed as the force times the component of the displacement in the direction of the force. Since \( \theta\) is the angle between the direction of the normal force and the displacement, the displacement in the direction of the normal force is zero.
04

Conclusion

Since the displacement component in the direction of the normal reaction force is zero, the work done by the normal reaction force must be zero.

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Key Concepts

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

normal reaction force
The normal reaction force, often known as the normal force, is a crucial concept. It is the force exerted by a surface, perpendicular to the object in contact with it.
In the context of an inclined plane, the normal reaction force acts perpendicular to the surface of the plane.
If you imagine a particle placed on an inclined plane, gravity pulls it down vertically while the inclined plane exerts a normal force perpendicular to its surface.
This force ensures the particle does not pass through the plane, balancing the vertical component of gravity.
inclined plane
An inclined plane is a flat surface tilted at an angle to the horizontal. This simple machine allows studying various forces and motions. When a particle or an object is placed on it, several forces act on it: gravitational force, normal reaction force, and frictional force (if friction is present).
Gravity can be resolved into two components on an inclined plane:
  • Parallel to the plane: This pulls the object down the slope.
  • Perpendicular to the plane: Balanced by the normal reaction force.

Understanding how these components interact helps analyze movements and energy transitions on inclined surfaces.
work and energy
Work and energy are essential concepts in physics. Work is defined as the transfer of energy by a force causing displacement.
Mathematically, work is represented as:

\( W = \textbf{F} \bullet \textbf{d} \)
Here,
Work depends on:
  • The magnitude of the force applied.
  • The displacement caused by the force.
  • The angle between the force and displacement directions.
In the context of an inclined plane, the work done by the normal reaction force on a particle is zero.
This is because the displacement in the direction of the normal force is zero (as outlined in the solution).
dot product
The dot product is a mathematical operation that takes two vectors and returns a scalar.
It is instrumental in calculating work done by forces.
The dot product of two vectors is given by:
\( \textbf{F} \boldsymbol{\bullet} \textbf{d} = |\textbf{F}| |\textbf{d}| \text{cos}(\theta) \)
where:
  • \( |\textbf{F}| \) is the magnitude of the force.
  • \( |\textbf{d}| \) is the magnitude of the displacement.
  • \( \theta \) is the angle between the force and displacement directions.

In the provided exercise, since the normal reaction force is perpendicular to the displacement (\( \theta = 90^\text{o} \)), the cosine of 90 degrees is zero.
Hence, the work done (dot product) by the normal reaction force is zero.

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Most popular questions from this chapter

A light elastic string, of unstretched length \(a\) and modulus of elasticity \(W\), is fixed at one end to a point on the ceiling of a room. To the other end of the string. is attached a particle of weight \(W\). A horizontal force \(P\) is applied to the particle and in equilibrium it is found that the string is stretched to three times its natural length. Calculate: (a) the angle the string makes with the horizontal, (b) the value of \(P\) in terms of \(W\). If, instead, \(P\) is not applied horizontally find the least value of \(P\) which in equilibrium will make the string have the same inclination to the horizontal as before. Deduce that the stretch length of the string is \(\frac{3}{2} a\) in this case and find the inclination of \(P\) to the vertical. (U of L)

A body of mass \(2.5\) kilogramme is attached to the end \(B\) of a light elastic string AB of natural length 2 metre and modulus \(5 g\) newton. The mass is suspended vertically in equitibrium by the string whose other end \(A\) is attached to a fixed point. (i) Find the depth below A of B when the body is in equilibrium. (ii) Find the distance through which the body must be pulled down vertically from its equilibrium position so that it will just reach A after release.

A mass of 3 kilogramme is connected by an elastic string of natural length 1 metre and modulus of elasticity \(14.7 \mathrm{~N}\) to a fixed point. A horizontal force equal to the weight of 1 kilogramme acts on the mass maintaining it in equilibrium. Find the inclination of the string to the vertical. If the horizontal force is removed, what is the least force which must act on the particle to ensure that the. string shall be inclined at the same angle as before. Calculate in each case the extension of the string.

The potential energy of a body of mass \(m\) is \(m g h\) where \(h\) is: (a) the distance from a chosen level, (b) the height above the ground, (c) the height above a chosen level, (d) the vertical distance moved.

A particle of weight \(W\) is attached by two light inextensible strings cach of length \(a\) to two fixed points distant \(a\) apart in a horizontal line. Write down the tension in either string. One of the strings is now replaced by an elastic string of the same natural length, and it is found that in the new position of equilibrium this string has stretched to a length \(5 a / 4\). Prove that the modulus of elasticity of this string is \(7 W / \sqrt{39}\), and show that the tension in the other string has been increased in the ratio \(5: \sqrt{13}\). (U of L)
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