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(C) The force is given by $F = -20x + 10 = -20(x - 0.5)$. Let $X = x - 0.5$. Then $F = -20X$. This represents a Simple Harmonic Motion $(SHM)$ about t

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$0.5 \ m, -10 \ kg \ m/s$

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(C) The force is given by $F = -20x + 10 = -20(x - 0.5)$. Let $X = x - 0.5$. Then $F = -20X$. This represents a Simple Harmonic Motion $(SHM)$ about the equilibrium position $x_0 = 0.5 \ m$.Comparing with $F = -m\omega^2 X$,we have $m\omega^2 = 20$. With $m = 5 \ kg$,$\omega^2 = 4$,so $\omega = 2 \ rad/s$.The amplitude $A$ is the distance from the equilibrium position to the starting point. At $t = 0$,$x = 1 \ m$,so $A = |1 - 0.5| = 0.5 \ m$.The equation of motion is $X(t) = A \cos(\omega t + \phi)$. At $t = 0$,$X = 0.5$,so $0.5 = 0.5 \cos(\phi) \Rightarrow \phi = 0$.Thus,$x(t) = 0.5 + 0.5 \cos(2t)$.At $t = \pi/4 \ s$,$x = 0.5 + 0.5 \cos(2 \cdot \pi/4) = 0.5 + 0.5 \cos(\pi/2) = 0.5 + 0 = 0.5 \ m$.The velocity is $v = dx/dt = -0.5 \cdot 2 \sin(2t) = -1 \sin(2t)$.At $t = \pi/4 \ s$,$v = -1 \sin(\pi/2) = -1 \ m/s$.Momentum $p = mv = 5 	imes (-1) = -5 \ kg \ m/s$. (Note: Correcting the calculation,the momentum is $-5 \ kg \ m/s$ at $x=0.5 \ m$).

$A$ block of mass $5 \ kg$ moves along the $x$-direction subject to the force $F = (-20x + 10) \ N$,with the value of $x$ in metre. At time $t = 0 \ s$,it is at rest at position $x = 1 \ m$. The position and momentum of the block at $t = (\pi / 4) \ s$ are

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