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Chapter 5: Problem 46

Which of the following are correct? (a) \(\Lambda\) body of wcight \(1 \mathrm{~N}\) has a potential energy of \(1 \mathrm{~J}\) relative to the ground when it is at a hcight of \(1 \mathrm{~m}\). (b) \(\Lambda \mathrm{l} \mathrm{kg}\) body has a kinetic energy of \(1 \mathrm{~J}\) when its velocity is \(1.14 \mathrm{lms}^{-1}\). (c) The power of an agent is \(\vec{F} \cdot \vec{v}\). (d) When a force retards the motion of a body, the work done is negative.

Short Answer

Expert verified
Statements (a), (c), and (d) are correct.

Step by step solution

01

Analyze Statement (a)

The potential energy of an object relative to the ground is calculated by \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 m/s虏), and \( h \) is the height. Given that \( PE = 1\, \text{J} \), \( h = 1\, \text{m} \), and \( W = 1\, \text{N} = mg \), let's verify if it satisfies \( PE = mgh \). Using \( mg = W = 1\, \text{N} \) and \( h = 1\, \text{m} \), \( PE = 1 \times 1 = 1 \) J. Thus, this statement is correct.
02

Analyze Statement (b)

The kinetic energy of a body is calculated using the formula \( KE = \frac{1}{2}mv^2 \). Here, the velocity \( v = 1.14\, \text{m/s} \) and \( KE = 1\, \text{J} \). Solving for mass \( m \), we have \( 1 = \frac{1}{2}m \times (1.14)^2 \), which gives \( m = \frac{2}{(1.14)^2} \approx 1.54\, \text{kg} \). Since \( m \) is not 1 kg, statement (b) is incorrect.
03

Analyze Statement (c)

The power of an agent is defined as the rate of doing work, often expressed as \( P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} \), where \( \vec{F} \) is the force and \( \vec{v} \) is the velocity vector. This is the definition of instantaneous power, making statement (c) correct.
04

Analyze Statement (d)

Work done by a force is calculated as \( W = \vec{F} \cdot \vec{s} = F \cdot s \cdot \cos\theta \), where \( \theta \) is the angle between the force and direction of displacement. If the force retards motion, \( \theta > 90^\circ \), and the work done \( W \) is negative because \( \cos\theta \) is negative. Hence, statement (d) is correct.

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Key Concepts

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

Potential Energy
Potential energy refers to the energy that an object possesses due to its position or state. Imagine an object positioned at a height from the ground; it has the potential to do work due to the energy stored within it. This potential energy is calculated through the formula: \[ PE = mgh \] Here:
  • \( m \) is the object鈥檚 mass (in kilograms)
  • \( g \) is the gravitational pull (\( 9.8\, \text{m/s}^2 \) on Earth)
  • \( h \) is the height from which the object is elevated
Real-life examples include a book placed on a shelf. If the shelf is 2 meters high, the book has stored potential energy. If it falls, that energy is converted to make the book move. It's a simple yet profound concept that's essential for understanding mechanics.
Kinetic Energy
Kinetic energy is the energy of motion. When an object moves, it carries kinetic energy with it. It's calculated using the equation: \[ KE = \frac{1}{2}mv^2 \] Where:
  • \( m \) is the mass of the object
  • \( v \) is the velocity (speed with direction) of the object
For instance, if a cyclist is moving down a road, their mass and speed determine their kinetic energy. Faster speeds lead to higher kinetic energy, making this concept vital for understanding how motion and energy interrelate. This formula helps illustrate why stopping a fast-moving car requires more effort 鈥� it has significant kinetic energy to be reduced to rest.
Power
Power is all about the rate at which work is done. Picture someone climbing stairs; power measures how quickly they reach the top. In physics, power is defined by: \[ P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} \] This formula means that power equals the rate of work done (\( \,dW/dt \, \)) or, equivalently, the dot product of force (\( \vec{F} \)) and velocity (\( \vec{v} \)).
  • Higher power indicates more work done in less time.
  • It鈥檚 measured in watts (joules per second).
This concept is crucial not only in physics but also in understanding machinery, engines, and even daily activities like running and lifting.
Work Done
Work done is the measure of energy transfer when a force moves an object. If you've ever pushed a toy car across a room, you've done work. It's computed as: \[ W = \vec{F} \cdot \vec{s} = F \cdot s \cdot \cos\theta \] Where:
  • \( \vec{F} \) signifies force applied
  • \( \vec{s} \) stands for displacement
  • \( \theta \) is the angle between force and movement direction
The result can be positive or negative, depending on the force direction. For example, pulling a box upwards does positive work, while slowing it down as it descends means doing negative work. This understanding is essential in both everyday scenarios and complex systems, serving as a cornerstone of mechanics.

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Most popular questions from this chapter

A block of mass \(m\) moves on a horizontal circle against the wall of a cylindrical room of radius \(R\). The lloor of the room on which the block moves in smooth but the friction coefficient between the wall and the block is \(\mu\). The block is given an initial speed \(v_{0} .\) Find the power developed by the resultant force acting on the block as a function of distance travelled \(s\).

In certain position \(X\), kinetic energy of a particle is \(25 \mathrm{~J}\) and potential energy is \(-10 \mathrm{~J} .\) In another position \(Y\), kinetic energy of the particle is \(95 \mathrm{~J}\) and the potential energy is \(-25 \mathrm{~J}\). During the displacement of the particle from \(X\) to \(Y\) (a) Net work done by all the forces is \(-70 \mathrm{~J}\). (b) Work done by the conservative forces is \(-15 \mathrm{~J}\). (c) Work done by all the forces besides the conservative forces is \(55 \mathrm{~J}\). (d) Work done by the conservative forces equals work done by the non- conservative forces.

Column-I (a) Force is cqual to (b) For the conscrvative force (c) Power is cqual to (d) Arca of \(P\) -t curve gives Column-II (p) Work is path independent (q) Rate of change of lincar momentum (r) Rate of work done (s) Produet of corce to the velocily (l) Work donc (u) Negative of the powntial cnergy gradicnt

A \(0.1 \mathrm{~kg}\) block is pressed against a horizontal spring fixed at one end to compress the spring through \(5 \mathrm{~cm}\). The spring constant is \(100 \mathrm{~N} / \mathrm{m}\). The ground is \(2 \mathrm{~m}\) below the spring. Which of the following is/are correct? (a) When released, the block shall have a kinetic energy of \(\frac{1}{8} \mathrm{~J}\). (b) The initial hoizontal velocity of the block is \(\sqrt{\frac{5}{2}} \mathrm{~m} / \mathrm{s}\). (c) The block shall reach the ground in \(\sqrt{\frac{2}{5}} \sec\). (d) The block will hit the ground at a horizontal distance of 1 metre from the free end of the spring.

\(\Lambda\) single conservative force \(F\) acts on \(1 \mathrm{~kg}\) particle that moves along \(x\) -axis. The potential encrgy \(U\) is given by \(U=20+(x-2)^{2}\) where \(x\) is in meters. \(\Lambda t x=5 \mathrm{~m}\) the particle has a kinetic energy of \(20 \mathrm{~J}\). 1\. What is mechanical energy of the system (a) \(29 \mathrm{~J}\) (b) \(20 \mathrm{~J}\) (c) \(49 \mathrm{~J}\) (d) \(19 \mathrm{~J}\) 2\. The maximum kinetic cnergy of particle (a) \(29 \mathrm{~J}\) (b) \(20 \mathrm{~J}\) (c) \(49 \mathrm{~J}\) (d) \(98 \mathrm{~J}\) 3\. For what (finitc) value of \(x\), does \(F(x)=0 ?\) (a) \(x=2 \mathrm{~m}\) (b) \(x=4 \mathrm{~m}\) (c) \(x=0 \mathrm{~m}\) (d) \(x=1 \mathrm{~m}\)
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