/*! 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 39 A body of mass \(2 \mathrm{~kg}\... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 39

A body of mass \(2 \mathrm{~kg}\) is moved from a point \(A\) to a point \(B\) by an external agent in a conservative force field. If the velocity of the body at the points \(A\) and \(B\) are \(5 \mathrm{~m} / \mathrm{s}\) and \(3 \mathrm{~m} / \mathrm{s}\), respectively, and the work done by the external agent is \(-10 \mathrm{~J}\), then the change in potential energy between points \(A\) and \(B\) is (A) \(6 \mathrm{~J}\) (B) \(36 \mathrm{~J}\) (C) \(16 \mathrm{~J}\) (D) None of these

Short Answer

Expert verified
The change in potential energy between points A and B is \(6\;\text{J}\). The correct answer is (A) \(6\;\text{J}\).

Step by step solution

01

Calculate the kinetic energy at points A and B

First, let's calculate the kinetic energy of the body at points A and B. The kinetic energy is given by the formula: \(KE = \dfrac{1}{2} m v^2\) At point A, the body has a velocity of 5 m/s. Thus, its kinetic energy is: \(KE_A = \dfrac{1}{2} (2\;\text{kg}) (5\;\text{m/s})^2 = 25\;\text{J}\) At point B, the body has a velocity of 3 m/s. Thus, its kinetic energy is: \(KE_B = \dfrac{1}{2} (2\;\text{kg}) (3\;\text{m/s})^2 = 9\;\text{J}\)
02

Apply the Work-Energy Theorem

Now, let's apply the work-energy theorem to find the change in potential energy between points A and B. The work-energy theorem states that: Work done = Change in kinetic energy + Change in potential energy Given that the work done by the external agent is -10 J, we can write this as: \(-10\;\text{J} = (KE_B - KE_A) + \Delta PE\)
03

Find the change in potential energy

Solving for the change in potential energy, we get: \(\Delta PE = -10\;\text{J} - (KE_B - KE_A) = -10\;\text{J} - (9\;\text{J} - 25\;\text{J})\) \(\Delta PE = -10\;\text{J} + 16\;\text{J} = 6\;\text{J}\) The change in potential energy between points A and B is 6 J. The correct answer is (A) \(6\;\text{J}\).

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Most popular questions from this chapter

A block of mass \(m=0.1 \mathrm{~kg}\) is released from a height of \(4 \mathrm{~m}\) on a curved smooth surface. On the horizontal smooth surface, it collides with a spring of force constant \(800 \mathrm{~N} / \mathrm{m}\). The maximum compression in spring will be \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(1 \mathrm{~cm}\) (B) \(5 \mathrm{~cm}\) (C) \(10 \mathrm{~cm}\) (D) \(20 \mathrm{~cm}\)

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