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In a certain situation, \(\mathbf{F}\) and \(\vec{s}\) are not equal to zero but the work done is zero. From this, we conclude that: \(\rightarrow \quad \rightarrow\) (a) \(\mathbf{F}\) and \(\vec{s}\) are in same direction (b) \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{s}}\) are perpendicular to each other \(\Rightarrow\) (c) \(\mathrm{F}\) and \(\overrightarrow{\mathrm{s}}\) are in opposite direction (d) none of the above

Short Answer

Expert verified
Option (b): \(\mathbf{F}\) and \(\vec{s}\) are perpendicular to each other.

Step by step solution

01

Understand the Concept of Work

Work is defined as the product of force and displacement in the direction of the force. Mathematically, it is given by the dot product \(W = \mathbf{F} \cdot \vec{s} = |\mathbf{F}| |\vec{s}| \cos(\theta)\), where \(\theta\) is the angle between the force \(\mathbf{F}\) and the displacement \(\vec{s}\).
02

Analyze the Given Information

We are told that the work done is zero. According to the formula for work, \(W = |\mathbf{F}| |\vec{s}| \cos(\theta) = 0\). Since \(\mathbf{F}\) and \(\vec{s}\) are not zero, the only way for their product to be zero is if \(\cos(\theta) = 0\).
03

Determine the Angle Condition

The cosine of an angle is zero when the angle \(\theta\) is \(90^\circ\). This means that the force and displacement are perpendicular to each other.
04

Conclusion Based on the Angle Condition

Based on the angle condition where \(\theta = 90^\circ\), the respective condition matches option (b), which states that "\(\mathbf{F}\) and \(\vec{s}\) are perpendicular to each other."

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Key Concepts

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

Dot Product
The dot product, a fundamental concept in vector mathematics, measures the interaction between two vectors. It is denoted by a dot (\(\cdot\)) and is calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. In equation form, the dot product \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \). This operation shows how much of one vector goes in the direction of another. In the context of work and energy, when calculating work done, the dot product helps determine how effectively a force contributes to an object's movement. If the angle \( \theta \) is zero degrees, \( \cos(\theta) \) is 1, resulting in maximum work being done as the force is completely directed along the displacement. Conversely, if \( \theta \) is 90 degrees, \( \cos(\theta) \) is 0, meaning no work is being done despite the presence of force and displacement, as they are perpendicular.
  • Zero dot product: Vectors are perpendicular.
  • Positive dot product: Vectors are in the same general direction.
  • Negative dot product: Vectors are in opposite directions.
Recognizing these scenarios can aid in visualizing physical situations like the conditions where force applied does no work.
Force and Displacement
Force and displacement are the two main components involved in performing mechanical work. It's crucial to understand how they work together to accomplish tasks. Force is a vector quantity, meaning it has both magnitude and direction. It acts as a push or pull upon an object resulting from its interaction with another object. Displacement, also a vector, represents the change in position of an object. It has magnitude, which is the straight-line distance between initial and final positions, and direction, showing where the object moves. The relationship between force and displacement determines the work done, evaluated using the dot product of these vectors. Work is essentially the result of how a force causes an object to move. If force and displacement are aligned, all force contributes to transmission movement. However, if they are perpendicular, as explored in the original problem, none of the force aids in displacement, leading to zero work. Keep in mind:
  • Same direction: Maximum work achieved as the path and force coincide.
  • Opposite direction: Negative work where force opposes motion.
  • Perpendicular: Zero work due to no force component in the direction of displacement.
These variations highlight how the combination of force magnitude and its direction relative to displacement influences work.
Angle Conditions
Angle conditions play a vital role in determining the effectiveness of work done when force is applied. In vector mathematics, the angle between the force vector, \( \mathbf{F} \), and the displacement vector, \( \vec{s} \), significantly impacts the calculated work. Understanding these angle scenarios is crucial for predicting whether work done will be positive, negative, or zero. The angle \( \theta \) affects the cosine value, crucial for calculating work: \( W = |\mathbf{F}| |\vec{s}| \cos(\theta) \). - **Zero angle (0°):** Cosine is 1, indicating maximum work when force and displacement are in the same direction.- **Right angle (90°):** Cosine becomes 0, showing perpendicular components where no work is done since the force doesn't contribute to displacement.- **Obtuse angle (greater than 90°):** Cosine is negative, indicating negative work when force opposes movement to a certain extent.The original problem illustrates these principles when zero work implies \( \theta = 90^\circ \), aligning with perpendicular vectors. Angle conditions help us visualize practical scenarios, enabling predictions about how effective a force will be in producing desired motion.

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Most popular questions from this chapter

A body of mass \(m\) was slowly halved upon the hill by a force which at each point was directed along a tangent to the path. The work done by the applied force:(a) does not depend upon path followed by the body (b) depends upon path(c) does not depend upon position of \(A\) and \(B\) (d) both (a) and (c) are correct

A block of mass \(m\) slides from the rim of a hemispherical bowl of radius \(R\). The velocity of the block at the bottom will be : (a) \(\sqrt{R g}\) (b) \(\sqrt{2 R g}\) (c) \(\sqrt{2 \pi R g}\) (d) \(\sqrt{\pi R g}\)

A force \(\vec{F}=-k(y \hat{i}+x \hat{j})\) (where \(k\) is a positive constant) acts on a particle moving in the \(x-y\) plane starting from the origin, the particle is taken along the positive \(x\) -axis to the point \((a, 0)\) and the parallel to the \(y\) -axis to the point \((a, a)\). The total work done by the force \(\overrightarrow{\mathbf{F}}\) on the particle is:(a) \(-2 \mathrm{ka}^{2}\) (b) \(2 \mathrm{ka}^{2}\) (c) \(-\mathrm{ka}^{2}\) (d) \(\mathrm{ka}^{2}\)

A stone tied to a string of length \(L\) is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed \(u\). The magnitude of the change in its velocity as it reaches a position where the string is horizontal is : (a) \(\sqrt{u^{2}-2 g L}\) (b) \(\sqrt{2 g L}\) (c) \(\sqrt{u^{2}-g L}\) (d) \(\sqrt{2\left(u^{2}-g L\right)}\)

A particle moves along the \(x\) -axis from \(x=0\) to \(x=5 \mathrm{~m}\) under the influence of a force given by \(F=7-2 x+3 x^{2}\) The work dune in the process is: (a) \(70 \mathrm{~J}\) (b) \(270 \mathrm{~J}\) (c) \(35 \mathrm{~J}\) (d) \(135 \mathrm{~J}\)

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