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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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If the K.E. of a body is increased by \(44 \%\), its momentum will increase by....... (A) \(20 \%\) (B) \(22 \%\) (C) \(2 \%\) (D) \(120 \%\)

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