A single conservative force F(x) acts on a 1.0-kg particle that moves along the x-axis. The potential energy U(x) is given by U(x)=20+(x-2)² where x is in metres. At x=5.0m, the particle has a kinetic energy of 20J select the correct options.

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The maximum and minimum values of x, respectively are 7.38 m, -3.38m.

The maximum kinetic energy of the particle and the value of x at which maximum kinetic energy occurs are 29J, 0m.

equation of force F(x) as a function of x is F=2(2-x)

The mechanical energy of the system is 45J

answer is B, D.

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Detailed Solution

(a) At x=5m, U=20+ (s-2)²=29J, K=20J mechanical energy= E= U+K=49J

(b) U_max= $= 49 J \Rightarrow 20 + (X - 2)^{2} = 49 \Rightarrow (X - 2)^{2} = 29$

$\Rightarrow X - 2 = \pm \sqrt{29} \Rightarrow X = - 3.38 M, 7.38 m$

(c) The maximum value of kinetic energy occurs when the potential energy is minimum, U_min=20J so K_max=29J, this occurs at x=2m

d) $F (x) = \frac{- d U}{d x} = - 2 (x - 2) = 2 (2 - x)$

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