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Chapter 4: Problem 131

When a rubber band is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x+\) \(b x^{2}\), where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubber band by \(L\) is: [2014] (A) \(a L^{2}+b L^{3}\) (B) \(\frac{1}{2}\left(a L^{2}+b L^{3}\right)\) (C) \(\frac{a L^{2}}{2}+\frac{b L^{3}}{3}\) (D) \(\frac{1}{2}\left(\frac{2 L^{2}}{2}+\frac{b L^{2}}{3}\right)\)

Short Answer

Expert verified
The work done in stretching the unstretched rubber band by a distance L is: \[W = \frac{a L^2}{2} + \frac{b L^3}{3}\]

Step by step solution

01

Write down the formula for work done

Work done (W) is equal to the integral of the force (F) with respect to distance (x). In this case, the distance is from 0 to L: \[W = \int_{0}^{L} F dx\]
02

Substitute the given force equation

Substitute the given force equation, \(F = ax + bx^2\), into the integral: \[W = \int_{0}^{L} (ax + bx^2) dx\]
03

Calculate the integral

Calculate the integral of force with respect to x: \[W = \left[\frac{a x^2}{2} + \frac{b x^3}{3}\right]_0^L\]
04

Substitute the limits of integration

Now, substitute the limits of integration (0 and L) into the expression: \[W = \left(\frac{a L^2}{2} + \frac{b L^3}{3}\right) - \left(\frac{a\cdot 0^2}{2} + \frac{b\cdot 0^3}{3}\right)\]
05

Simplify the expression

Simplify the expression to obtain the final equation for work done: \[W = \frac{a L^2}{2} + \frac{b L^3}{3}\] The correct answer is (C).

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Key Concepts

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

Work Done Calculation
Understanding the concept of work done is crucial when solving physics problems related to force and energy. Work done is to define the amount of energy transferred by a force acting over a distance. In the context of the JEE Main Physics problem, the work done on a rubber band as it is stretched involves calculating the total energy used to move the rubber band from its original position to a stretched position of length L.

The work done by a varying force is not as straightforward as multiplying force by distance, because the force changes with the distance, as seen with the rubber band. The formula for work done by a variable force is given by the integral of force with respect to distance. In simpler terms, we sum up the tiny bits of work done over each infinitesimal distance covered to get the total work done. The correct approach, as shown in the step-by-step solution, involves setting up an integral that accumulates the work done across the stretch of the rubber band.
Elastic Potential Energy
The elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Elastic potential energy is equal to the work done to stretch or compress the object. In the case of the rubber band problem, this energy is what gets stored as the band is stretched and can be recovered when the band is released, propelling it back to its original shape.

The work done in stretching the rubber band gives rise to its potential energy, which, according to Hooke's law, is dependent on the displacement squared when dealing with a perfect spring. However, for the rubber band which follows a non-linear force law, the relationship includes higher powers of displacement, and the associated potential energy is calculated through integration. Calculating the work done as shown in the solution properly provides us with the amount of elastic potential energy stored in the rubber band upon being stretched by the distance L.
Integration in Physics
Integration is widely used in physics to solve problems where quantities vary continuously. It is a mathematical tool that enables us to find quantities like area, volume, and, as in our problem, work done when a force varies with respect to distance. The process of integration considers the sum of infinite infinitesimally small quantities to find a total amount.

In the context of the problem from the JEE Main, integration is used to calculate the work done by a force that changes as the rubber band is stretched. The formula for the force includes terms for both linear (\( ax \) and quadratic (\( bx^2 \) variation with distance and hence their integration does not result in a single power of x but rather a combination that reflects the work done for each contribution. By integrating and evaluating the function from the initial to the final point of stretching, we can precisely ascertain the work done in stretching the rubber band.

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Most popular questions from this chapter

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle; the motion of the particle takes place in a plane. It follows that (A) Its kinetic energy is constant. (B) Its acceleration is constant. (C) Its velocity is constant. (D) It moves in a straight line.

A block of mass \(2 \mathrm{~kg}\) is lifted through a chain. When block moves through \(2 \mathrm{~m}\) vertically the velocity becomes \(4 \mathrm{~m} / \mathrm{s}\). Work done by chain force until it moves \(2 \mathrm{~m}\) is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (A) \(40 \mathrm{~J}\) (B) \(24 \mathrm{~J}\) (C) \(56 \mathrm{~J}\) (D) None of these

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 body of mass \(m\) accelerates uniformly from rest to \(v_{1}\) in time \(t_{1} .\) The instantaneous power delivered to the body as a function of time \(t\) is [2004] (A) \(\frac{m v_{1} t^{2}}{t_{1}}\) (B) \(\frac{m v_{1}^{2} t}{t_{1}^{2}}\) (C) \(\frac{m v_{1} t}{t_{1}}\) (D) \(\frac{m v_{1}^{2} t}{t_{1}}\)

A block of mass \(m\) is released from rest when the extension in the spring is \(x_{0}\). The maximum downward displacement of the block is (A) \(\frac{m g}{2 k}-x_{0}\) (B) \(\frac{m g}{2 k}+x_{0}\) (C) \(\frac{2 m g}{k}-x_{0}\) (D) \(\frac{2 m g}{k}+x_{0}\)
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