Answer
See the detailed answer below.
Work Step by Step
a) We know that the position of an object that undergoes a simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)$$
Now we can use the conservation of energy law since the mass is oscillating without friction.
$$\color{red}{\bf\not}\frac{1}{2}kA^2=\color{red}{\bf\not}\frac{1}{2}mv^2+\color{red}{\bf\not}\frac{1}{2}kx^2 $$
Hence,
$$ A=\sqrt{\dfrac{ mv^2+ kx^2 }{k}}$$
$$ A=\sqrt{\dfrac{ mv^2}{k}+ x^2 }$$
So, at $t=0$, $x=-5$ cm, and $v=20$ cm/s.
Now we need to plug the known;
$$ A=\sqrt{\dfrac{ (0.1)(0.20)^2}{2.5}+ (-0.05)^2 }=\bf 0.064$$
$$A=\color{red}{\bf 6.4}\;\rm cm$$
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b) The block maximum acceleration is given by
$$a_{\rm max}=A\omega^2$$
where $\sqrt{\dfrac{k}{m}}$
$$a_{\rm max}=A\left( \dfrac{k}{m}\right)$$
Plugging the known;
$$a_{\rm max}=(6.4\times 10^{-2})\left( \dfrac{2.5}{0.1}\right)$$
$$a_{\rm max}=\color{red}{\bf1.6}\;\rm m/s^2$$
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c) We need to find the position at which the acceleration is at its maximum value. This occurs when the block is at the most stretched point away from the equilibrium point.
So when $a=a_{\rm max}$, $x=-A$.
The negative sign here is due to the direction of the acceleration since we need it in the positive direction, the position of the object must be in the negative direction.
$$x_{a_{\rm max}}=\color{red}{\bf- 6.4}\;\rm cm$$
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d) To find the speed of the block when $x=3$ cm, we can use the conservation of energy.
$$ \color{red}{\bf\not}\frac{1}{2}mv_0^2+ \color{red}{\bf\not}\frac{1}{2}kx_0^2= \color{red}{\bf\not}\frac{1}{2}mv_1^2+ \color{red}{\bf\not}\frac{1}{2}kx_1^2$$
$$ mv_0^2+ kx_0^2= mv_1^2+ kx_1^2$$
$$v_1^2=\dfrac{mv_0^2+ kx_0^2-kx_1^2}{m}$$
$$v_1 =\sqrt{v_0^2+\dfrac{ k(x_0^2-x_1^2)}{m}}$$
Plugging the known;
$$v_1 =\sqrt{(0.20)^2+\dfrac{ 2.5[(-0.05)^2-(0.03)^2]}{0.1}}=\bf 0.283\;\rm m/s$$
$$v_1=\color{red}{\bf 28.3}\;\rm m/s$$
