$|\vec{R}|=|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2 A B \cos \theta}$
$|\vec{A}|=|\vec{B}|=|\vec{R}|=1$
$1=1+1+2 \times 1 \times 1 \times \cos \theta$
$\cos \theta=-\frac{1}{2} \Rightarrow \theta=120^{\circ}$
$|\vec{R}|=|\vec{A}-\vec{B}|=\sqrt{A^2+B^2-2 A B \cos 120^{\circ}}$
$=\sqrt{1^2+1^2-2 \times 1 \times 1 \times\left(-\frac{1}{2}\right)}=\sqrt{3}=|\vec{A}-\vec{B}|$
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$A$: standing on the horizontal surface
$B$: standing on the block
According to observer $B$, the potential energy of the spring increases
