/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 A position-dependent force \(F=x... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 29

A position-dependent force \(F=x^{2}-3\) Newton acts on a small body of mass \(2 \mathrm{~kg}\) and displaces it from \(x=0\) to \(x=5 \mathrm{~m}\). The work done is (A) \(110 \mathrm{~J}\) (B) \(\frac{80}{3} \mathrm{~J}\) (D) \(\frac{95}{2} \mathrm{~J}\) (D) Zero

Short Answer

Expert verified
The work done is \(\frac{80}{3} J\). The correct option is (B).

Step by step solution

01

1. Identify the given parameters and required output

: Force, F = x^2 - 3 N Mass of the body, m = 2 kg Initial position, x = 0 Final position, x = 5 m We need to find the work done (W).
02

2. Write down the formula for work done

: We know that the work done by a position-dependent force F(x) acting on a body when it is displaced from position x=a to position x=b is given by the formula: W = \(\int_{a}^{b} F(x)dx\)
03

3. Substitute the given values and integrate

: In this case, we need to find the work done when the body is displaced from x=0 to x=5, i.e., we need to find W = \(\int_{0}^{5} (x^2 - 3)dx\). Now, let's integrate the expression to find the work done: W = \(\int_{0}^{5} (x^2 - 3)dx\) = \(\left[\frac{x^3}{3} - 3x\right]_0^5\)
04

4. Evaluate the integral at the limits and find the work done

: Now, let's evaluate the expression at the limits x=5 and x=0: W = \(\left[\frac{(5)^3}{3} - 3(5)\right] - \left[\frac{(0)^3}{3} - 3(0)\right]\) W = \(\frac{125}{3} - 15\) Now, we simplify and find the numerical value of W: W = \(\frac{125}{3} - \frac{45}{3}\), since we need a common denominator to subtract. W = \(\frac{80}{3}\) Hence, the work done is \(\frac{80}{3} J\). The correct option is (B).

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Most popular questions from this chapter

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface varies with its speed \(v\) as \(\mu=\frac{1}{\sqrt{1+v}} .\) The acceleration of the block when its speed is \(3 \mathrm{~m} / \mathrm{s}\) will be (A) \(\frac{P}{3 m}-\frac{g}{2}\) (B) \(\frac{P}{3 m}+\frac{g}{2}\) (C) \(\frac{P}{3 m}\) (D) \(\frac{g}{2}\)

An athlete in the Olympic Games covers a distance of \(100 \mathrm{~m}\) in \(10 \mathrm{~s}\). His kinetic energy can be estimated to be in the range [2008] (A) \(200 \mathrm{~J}-500 \mathrm{~J}\) (B) \(2 \times 10^{5} \mathrm{~J}-3 \times 10^{5} \mathrm{~J}\) (C) \(20,000 \mathrm{~J}-50,000 \mathrm{~J}\) (D) \(2,000 \mathrm{~J}-5,000 \mathrm{~J}\)

A position-dependent force \(F=7-2 x+3 x^{2} N\) acts on a small body of mass \(2 \mathrm{~kg}\) and displaces it from \(x=0\) to \(x=5 \mathrm{~m} .\) The work done in joule is (A) \(70 \mathrm{~J}\) (B) \(270 \mathrm{~J}\) (C) \(35 \mathrm{~J}\) (D) \(135 \mathrm{~J}\)

A block of mass \(m\) is slowly pulled up on inclined plane of height \(h\) and inclination \(\theta\) with the top of a string parallel to the incline. Which of the following statement is correct for the block when it moves up from the bottom to the top of the incline? (A) Work done by the normal reaction force is zero. (B) Work done by the string on block is \(m g h\). (C) Work done by the gravity is \(m g h\). (D) Work done by the block is \(-m g h / 2\).

A block of mass \(2 \mathrm{~kg}\) is held over a vertical spring with spring unstretched. Suddenly, if block is left free, maximum compression of spring is [spring constant \(K=200 \mathrm{~N} / \mathrm{m}]:\) (A) \(0.2 \mathrm{~m}\) (B) \(0.1 \mathrm{~m}\) (C) \(0.4 \mathrm{~m}\) (D) \(0.05 \mathrm{~m}\)
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