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Chapter 4: Problem 111

A homogeneous rope of mass per unit length \(\lambda\) and length \(l\) kept on ground and one end of the rope is fixed to ground at \(O .\) The left end of the rope (with respect to fixed end) is pulled by an external agent which imparts constant velocity \(v\) to it. Find the work done by the external agent (in joule) to place the moving end extremely right with respect to fixed end. Take \(\lambda=1 \mathrm{~kg} / \mathrm{m}, v=1 \mathrm{~ms}^{-1}\) and \(l=1 \mathrm{~m}\).

Short Answer

Expert verified
The work done by the external agent to place the moving end extremely right with respect to the fixed end is \(\frac{1}{2}\) Joule.

Step by step solution

01

Identify the variables given in the problem

Length of the rope, \(l=1 \mathrm{~m}\) Mass per unit length, \(\lambda=1 \mathrm{~kg} / \mathrm{m}\) Velocity by which external agent pulls the rope, \(v=1 \mathrm{~ms}^{-1}\)
02

Consider a small segment of the rope

Let's consider a small segment of the rope with length \(\mathrm{d} x\). The mass of the segment is \(\mathrm{d} m = \lambda \times \mathrm{d} x\).
03

Calculate the kinetic energy of the segment

The segment is moving with the velocity \(v\). So the kinetic energy of the segment is given by \[\mathrm{d} K = \frac{1}{2}\mathrm{d} m \times v^2 = \frac{1}{2} \lambda \mathrm{d} x \times v^2\].
04

Integrate kinetic energy over the entire length of the rope

Since each segment of the rope gains kinetic energy by the external agent, we need to integrate the kinetic energy over the entire length of the rope to find the total kinetic energy (work done) by the external agent. \[W = \int_0^l \frac{1}{2} \lambda v^2 \mathrm{d} x\].
05

Evaluate the integral

Now, we need to evaluate the integral to find the work done by the external agent. \[W = \frac{1}{2} \lambda v^2 \int_0^l \mathrm{d} x\]. Since the integral \(\int_0^l \mathrm{d} x = l\), we have: \[W = \frac{1}{2} \times 1 \times 1^2 \times 1 = \frac{1}{2} \,\mathrm{J}\]. So, the work done by the external agent to place the moving end extremely right with respect to the fixed end is \(\frac{1}{2}\) Joule.

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Key Concepts

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

Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. It is a fundamental concept in physics, encapsulating how an object鈥檚 velocity translates into energy.

Kinetic energy (\textbf{KE}) is given by the formula: \[KE = \frac{1}{2}mv^2\]where \textbf{m} is the mass of the object and \textbf{v} is its velocity. This relationship tells us that kinetic energy increases with the square of an object鈥檚 speed, which suggests a dramatic rise in energy with increasing velocity.

In the exercise using the rope, we calculate the kinetic energy for a small segment of the rope to understand the work done in motion. Incrementing this energy across the entire rope length through integration gives the total work needed to accelerate the rope from rest to the final velocity by the external agent.
Integration in Physics
Integration is a mathematical tool that is used extensively in physics to combine small quantities to find a total or aggregate quantity. When dealing with continuous variables, like a rope鈥檚 continuous mass distribution, integration allows us to sum up infinitesimally small segments smoothly.

In our example, the kinetic energy for a tiny segment of the rope is calculated first. Then, to find the total energy associated with the full length of the rope, we 鈥榠ntegrate鈥� these tiny energies over the length. The process of integration compiles the small energies \(\text{d}K\) over the entire length \(l\) of the rope:
\[W = \int_0^l \text{d}K\]The solution to this definite integral gives us the work done by the external agent. This demonstrates how integration can be a powerful tool for understanding physical processes and solving real-world problems.
Mass per Unit Length
The 'mass per unit length' (\(\textbf{lambda}, \lambda\)) is a property of an object that describes how its mass is distributed along its length. It鈥檚 especially relevant in continuous objects like strings or ropes.

For our homogeneous rope, this property implies that the mass is evenly distributed along its entire length. The mass per unit length is constant and can be used to find the mass of any segment of the rope simply by multiplying it by the length of the segment:\[dm = \lambda dx\]In physics problems, knowing the mass per unit length allows us to understand how the mass contributes to other physical quantities, such as kinetic energy. It is an essential element in problems involving extended objects where mass distribution affects motion, tension, or other dynamic properties. In our exercise, it is used to determine the kinetic energy of a segment and ultimately helps in calculating the total work done by the external force.

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Most popular questions from this chapter

A particle of mass \(5 \mathrm{~kg}\) moving in the \(X-Y\) plane has its potential energy given by \(U=(-7 x+24 y)\) Joule. The particle is initially at origin and has velocity \(\vec{u}=(14.4 \hat{i}+4.2 \hat{j}) \mathrm{m} / \mathrm{s}\) (A) The particle has speed \(25 \mathrm{~m} / \mathrm{s}\) at \(t=4 \mathrm{~s}\). (B) The particle has an acceleration \(5 \mathrm{~m} / \mathrm{s}^{2}\). (C) The acceleration of particle is normal to its initial velocity. (D) None of these.

A man squatting on the ground gets straight up and stands. The force of reaction of ground on the man during the process is (A) Constant and equal to \(\mathrm{mg}\) in magnitude. (B) Constant is greater than \(\mathrm{mg}\) in magnitude. (C) Variable but always greater than \(\mathrm{mg}\). (D) At first greater than \(\mathrm{mg}\) and later becomes equal to \(\mathrm{mg}\).

One end of an unstretched springs of force constant \(k_{1}\) is attached to the ceiling of an elevator. A block of mass \(1.5 \mathrm{~kg}\) is attached to other end. Another spring of force constant \(k_{2}\) is attached to the bottom of the mass and to the floor of the elevator as shown in Fig. \(4.28\). At equilibrium, the deformation in both the spring are equal and is \(40 \mathrm{~cm}\). If the elevator moves with constant acceleration upward, the additional deformation in both the springs have \(8 \mathrm{~cm} .\) Find the elevator's acceleration \(\left(g=10 \mathrm{~ms}^{-2}\right)\)

A cricket ball is hit for a six leaving the bat at an angle of \(45^{\circ}\) to the horizontal with kinetic energy \(K\). At the top position, the kinetic energy of the ball is (A) Zero (B) \(K\) (C) \(K / 2\) (D) \(K / \sqrt{2}\)

We generally ignore the kinetic energy of the moving coil of a spring but consider a spring of mass \(m\), equilibrium length \(L\) and spring constant \(k\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v .\) Assume that the speed of points along the length of the spring varies linearly with distance \(L\) from the fixed end. Assume also that the mass \(m\) of the spring is distributed uniformly along the length of the spring. Assume further that the force applied by the spring is spring constant times its deformation. In a spring gun, such a spring of mass \(0.243 \mathrm{~kg}\) and force constant \(3200 \mathrm{~N} / \mathrm{m}\) is compressed \(2.50 \mathrm{~cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizontally the ball of mass of \(0.053 \mathrm{~kg}\). The speed of small length \((d x)\) at a distance \(x\) from fixed end is (A) \(\frac{x}{L} v\) (B) \(v\) (C) \(\frac{L}{x} v\) (D) \(x v\)
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