Answer
a) $1.46 \;\rm m$
b) $19.6\;\rm cm$
Work Step by Step
a) Since the ramp is frictionless, we can use the conservation of energy law.
$$E_i=E_f$$
where $E_i$ is when the block is at its initial height before sliding down and $E_f$ is when the block hits the spring and stops.
$$K_i+U_{ig}+U_{is}=K_f+U_{fg}+U_{fs}$$
$$\frac{1}{2}mv_i^2+mgy_i+\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2+mgy_f+\frac{1}{2}kx_f^2$$
The initial velocity of the block is zero since it slides from rest and the final velocity is also zero since it finally stops after compressing the spring. The spring initially was not compressed or stretched, so $x_i=0$ as well.
$$\frac{1}{2}m(0)^2+mgy_i+\frac{1}{2}k(0)^2=\frac{1}{2}m(0)^2+mgy_f+\frac{1}{2}kx_f^2$$
We choose the final height to be the origin at which $y=0$, so
$$ mgy_i = mg(0)+\frac{1}{2}kx_f^2$$
$$ mgy_i = \frac{1}{2}kx_f^2$$
Solving for $x_f$;
$$x_f^2= \dfrac{2mgy_i}{k} $$
Now we need to find $y_i$. From the geometry of the right triangle in the figure below, we can see that
$$\sin\theta=\dfrac{y_i}{d+x_f}$$
where $d+x_f$ is the distance traveled by the block down the ramp.
Thus,
$$y_i=(d+x_f)\sin\theta$$
Plugging into (1);
$$x_f^2= \dfrac{2mg(d+x_f)\sin\theta}{k} $$
$$kx_f^2= 2mg\sin\theta(d+x_f) $$
$$kx_f^2= 2mg\sin\theta d+ 2mg\sin\theta x_f $$
Thus,
$$kx_f^2-2mg\sin\theta x_f - 2mgd\sin\theta =0 $$
Plugging the known;
$$250x_f^2-(2\times 10\times 9.8\sin30^\circ)x_f- (2\times 10\times 9.8\times 4\sin30^\circ)=0 $$
$$250x_f^2-98 x_f- 392=0 $$
Hence,
$x_f=\bf -1.0714\;\rm m$ or $x_f=\bf 1.4634\;\rm m$
Therefore, the maximum compression is
$$x_{f}=\color{red}{\bf 1.46}\;\rm m$$
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b) To find at what compression the speed of the box is maximized, we need to use the conservation of energy again.
$$K_i+U_{ig}+U_{is}=K_f+U_{fg}+U_{fs}$$
$$\frac{1}{2}mv_i^2+mgy_i+\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2+mgy_f+\frac{1}{2}kx^2$$
The initial velocity of the block is zero since it slides from rest but the final velocity is not zero, it is the maximum velocity. The spring initially was not compressed or stretched, so $x_i=0$ as well.
$$\frac{1}{2}m(0)^2+mgy_i+\frac{1}{2}k(0)^2=\frac{1}{2}mv^2+mgy_f+\frac{1}{2}kx^2$$
We choose the height at which the speed is maximum to be the origin at which $y=0$, so $y_f=0$.
Thus,
$$ mgy_i =\frac{1}{2}mv^2 +\frac{1}{2}kx^2$$
$$y_i=(d+x )\sin\theta$$
$$ mg(d+x )\sin\theta=\frac{1}{2}mv^2 +\frac{1}{2}kx^2$$
Now we need to find $v$ as a function of $x$ where $x$ is the compressed distance of the spring.
$$ mg d \sin\theta +mg (\sin\theta)x=\frac{1}{2}mv^2 +\frac{1}{2}kx^2$$
Multiplyin by 2;
$$ 2mg d \sin\theta +2mg (\sin\theta)x= mv^2 + kx^2$$
Solving for $v^2$;
$$ \dfrac{2mg d \sin\theta +2mg (\sin\theta)x- kx^2}{m}= v^2 $$
$$v^2= 2 g d \sin\theta +2 g (\sin\theta)x- \dfrac{kx^2}{m} $$
Taking the derivative with respect to $x$;
$$\dfrac{d}{dx}\left(v^2\right)= \dfrac{d}{dx}\left(2 g d \sin\theta +2 g (\sin\theta)x- \dfrac{kx^2}{m}\right) $$
$$2v\dfrac{dv}{dx} =0 +2 g (\sin\theta) - \dfrac{2kx }{m} $$
We know at the maximum value, $dv/dx=0$, so
$$2v(0)= 2 g \sin\theta - \dfrac{2kx }{m} $$
Thus,
$$ \color{red}{\bf\not}2 g \sin\theta = \dfrac{\color{red}{\bf\not}2kx }{m} $$
Solving for $x$;
$$ x=\dfrac{ mg \sin\theta}{ k} $$
Plugging the known;
$$ x=\dfrac{ 10\times9.8 \sin30^\circ}{ 250} =\bf 0.196\;\rm m $$
$$x=\color{red}{\bf 19.6}\;\rm cm$$
