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Chapter 4: Problem 490

A single conservative force \(\mathrm{F}(\mathrm{x})\) acts on a $2.5 \mathrm{~kg}\( particle that moves along the \)\mathrm{x}$ -axis. The potential energy \(\mathrm{U}(\mathrm{x})\) is given by \(\mathrm{U}(\mathrm{x})=\left[10+(\mathrm{x}-4)^{2}\right]\) where \(\mathrm{x}\) is in meter. \(\mathrm{At} \mathrm{x}=6.0 \mathrm{~m}\) the particle has kinetic energy of \(20 \mathrm{~J}\). what is the mechanical energy of the system? (A) \(34 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(48 \mathrm{~J}\) (D) \(49 \mathrm{~J}\)

Short Answer

Expert verified
= 10 + (2)^2 = 10 + 4 = 14 J #Step 2: Add the potential energy U to the given kinetic energy K to get the total mechanical energy E# #tag_title#Calculate the Total Mechanical Energy #tag_content#Mechanical Energy, E(x) = K + U E(6) = 20 J + 14 J = 34 J The mechanical energy of the system is \(34 J\). Answer: \(\boxed{(A)}\).

Step by step solution

01

Find the Potential Energy at x = 6.0 m

Given the potential energy function U(x) = [10 + (x - 4)^2], substitute x = 6.0 m into the function to calculate the potential energy at x = 6.0 m: U(6) = [10 +(6 - 4)^2]

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Most popular questions from this chapter

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

A bomb of mass \(10 \mathrm{~kg}\) explodes into 2 pieces of mass $4 \mathrm{~kg}\( and \)6 \mathrm{~kg}\(. The velocity of mass \)4 \mathrm{~kg}$ is \(1.5 \mathrm{~m} / \mathrm{s}\), the K.E. of mass \(6 \mathrm{~kg}\) is ....... (A) \(3.84 \mathrm{~J}\) (B) \(9.6 \mathrm{~J}\) (C) \(3.00 \mathrm{~J}\) (D) \(2.5 \mathrm{~J}\)

A body of mass \(\mathrm{m}\) is accelerated uniformly from rest to a speed \(\mathrm{v}\) in time \(\mathrm{T}\). The instantaneous power delivered to the body in terms of time is given by..... (A) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{\mathrm{T}^{2}\right\\}\right] \cdot \mathrm{t}$ (B) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{\mathrm{T}^{2}\right\\}\right] \cdot \mathrm{t}^{2}$ (C) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\\{2 \mathrm{~T}\\}\right] \cdot \mathrm{t}$ (D) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{2 \mathrm{~T}^{2}\right\\}\right] \cdot \mathrm{t}^{2}$

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(50 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(5 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table. $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{2}\right)$ (A) \(7.2 \mathrm{~J}\) (B) \(3 \mathrm{~J}\) (C) \(4.6 \mathrm{~J}\) (D) \(120 \mathrm{~J}\)
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