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Column-I (a) Work (b) Power (c) Force (d) Impulse Column-II (p) Slope of momentum \(p\) -t graph (q) Slope of \(W-t\) graph (r) Area under \(F-s\) graph (s) Area under \(F-f\) graph (t) Unit in SI system Joule

Short Answer

Expert verified
(a) - (r),(t); (b) - (q); (c) - (p); (d) - (s) (with typo consideration).

Step by step solution

01

Match Work with Column-II

Work, generally denoted by \(W\), is defined as the product of force and displacement. In graphical terms, this is represented by the area under the force-displacement graph (\(F-s\) graph). Thus, work corresponds to option \((r)\).
02

Match Power with Column-II

Power is defined as the rate of doing work. The slope of a work-time (\(W-t\)) graph represents power because it is the derivative of work with respect to time. Hence, power corresponds to option \((q)\).
03

Match Force with Column-II

Force is related to the change in momentum over time. The derivative of momentum with respect to time gives force, which is the slope of the momentum-time (\(p-t\)) graph. Thus, force corresponds to option \((p)\).
04

Match Impulse with Column-II

Impulse is the change in momentum experienced by an object and is equivalent to the integral of force over time, which is represented by the area under the force-time (\(F-t\)) graph. This should be option \((s)\), but since \((s)\) targets \(F-f\) graph (assuming \(F-t\) graph), it's a typo involving letter mix-up as common in exams, checking force-time relation and area fits conceptually.
05

Match Work with the SI Unit

The unit of work in the SI system is the Joule. Among the options, \((t)\) specifies this.

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Key Concepts

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

Work and Power concepts
Work and power are fundamental concepts in physics that describe how energy is transferred or transformed. **Work** is done when a force is applied to move an object over a distance. It is mathematically expressed as \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of movement.

In graphical representations, work can be visualized as the **area under the force-displacement (F-s) graph**. This means if you have a graph plotting force versus displacement, the area between the curve and the displacement axis gives the amount of work done.

**Power**, on the other hand, is the rate at which work is done. It is calculated by dividing work by time, \( P = \frac{W}{t} \). In terms of graphs, power corresponds to the **slope of the Work-Time (W-t) graph**. A steeper slope indicates higher power, meaning more work is done over a shorter period.
Force and Impulse analysis
Force and impulse are closely associated with changes in motion. **Force** is a vector that causes an object to accelerate and is defined as mass times acceleration (\( F = ma \)). It is related to the change in momentum, where the **slope of the Momentum-Time (p-t) graph** indicates the force since force is the derivative of momentum with respect to time.

**Impulse** represents the change in momentum and is given by the integral of force over time, \( I = \int F \cdot dt \). On a Force-Time graph, impulse is represented by the **area under the F-t graph**. Although there's often a mix-up with graph symbols, impulse captures how force applied over a period affects momentum growth or deceleration.
Graph interpretations in Physics
Graphs play a crucial role in physics as they can visually represent relationships between different quantities. Understanding how to interpret various physics graphs is essential for solving many problems.

For example, in a **Force-Displacement graph**, the **area under the curve** provides the work done. For a **Work-Time graph**, the **slope of the graph** indicates the power. Similarly, for a **Momentum-Time graph**, understanding the **slope** reveals the force acting on an object. In physical interpretations, where area and slope deduce different physical quantities, being able to switch between these views is vital for comprehensive understanding.

Always pay attention to the axes, units, and scales used in graphs to accurately interpret the data they communicate.
SI units in Physics
The International System of Units (SI units) is the standard unit system used in physics, ensuring consistency and uniformity worldwide. Understanding these units is key to solving physics problems accurately.

The SI unit of **work** is the Joule, which is equivalent to one Newton meter (1 Joule = 1 Nm). This relates to moving a force of one Newton over a distance of one meter. Similarly, the SI unit of **power** is the Watt, equivalent to one Joule per second (1 W = 1 J/s), indicating energy transfer per time unit.

A force is measured in Newtons, while impulse is expressed in Newton-seconds, derived from its relation to change in momentum. Familiarizing yourself with these units and how they relate to each other is crucial for mastering physics problems.

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Most popular questions from this chapter

No work is done by a force on an object if (a) the force is always perpendicular to its velocity. (b) the force is always perpendicular to its acceleration. (c) the object is stationary but the point of application of the force moves on the object. (d) the object moves in such a way that the point of application of the force remains fixed.

Statement-1: Work done by a force acting on a particle may be negative, even if kinetic energy of the body remains constant. Statement-2: Work done by conservative force equals negative of change in potential cnergy of Lhe body.

The potential energy of a \(4 \mathrm{~kg}\) particle free to move along the \(x\) -axis is given by \(U(x)-\frac{x^{3}}{3}-\frac{5 x^{2}}{2}+6 x+3\) Total mechanical cnergy of the particle is \(17 \mathrm{~J}\). Then the maximum kinctic cnergy is (a) \(10 \mathrm{~J}\) (b) \(2 \mathrm{~J}\) (c) \(9.5 \mathrm{~J}\) (d) \(0.5 \mathrm{~J}\)

A hemispherical bowl of radius \(R\) is rotating about its axis of symmetry which is kept vertical, A small ball kept in the bowl rotates with the bow! without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is \(\alpha\). Find the angular speed at which the bowl is rotating. Solution Lel \(\omega\) be the angular speed of rotation of the bowl two forces are acting on the ball. (a) normal reaction \(N\) (b) weight \(m g\) The ball is rolating in a circle of radius \(r(=R \sin \alpha)\) with centre at \(A\) at an angular specd \(\omega\). Thus, \(N \sin \alpha=m r \omega^{2}=m R \omega^{2} \sin \alpha \quad \ldots\) (i) \(\quad\) and \(\quad N \cos \alpha=m g \quad \ldots\) (ii) Dividing Eqs. (i) by (ii), we gel \(\frac{1}{\cos \alpha}-\frac{\omega^{2} R}{g}\) \(\omega-\sqrt{\frac{g}{R \cos \alpha}}\)

A heavy particle of weight \(w\), attached to a fixed point by a light inextensible string deseribes a circle in a vertical plane. The tension in the string has the values \(m w\) and \(n w\) respectively when the particle is at the highest and lowest points in the path. Then (a) \(m+n=6\) (b) \(\frac{m}{n}-2\) (c) \(m-n=-6\) (d) \(n-m=-6\)

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