A Galois deformation problem is said to be obstructed if we have a surjective map of Artinian rings $A_1 \to A_0$ (inducing the identity on residue fields $k$), with kernel $I$ such that $I\cdot m_{A_1} = 0$, and a Galois representation $\rho_0: G\to GL_N(A_0)$ that lacks a lift to $GL_N(A_1)$. This is equivalent to the non-triviality of the obstruction class $\mathcal{O}(\rho) \in H^2(\Pi, Ad(\overline{\rho_0})) \otimes I$, where $Ad(\overline{\rho})$ is a Galois module with action given by applying the residual representation $\overline{\rho}: G \to GL_N(k)$ and then conjugating. (This setup is taken from Barry Mazur's paper "Deforming Galois Representations.")
My question is, what is an example of such an obstruction?
Suppose that $G = \mathbb{Z}/p = \langle \tau \rangle$. Then the universal deformation ring of the trivial representation $G \rightarrow \mathrm{GL}_1(\mathbb{F}_p) = \mathbb{F}^{\times}_p$ is
$$R = \mathbb{Z}_p[[X]]/((1+X)^p - 1).$$
The universal deformation is given explicitly by $G \rightarrow R^{\times}$ sending $\tau$ to $1+X$. There is a map:
$$\rho: G \rightarrow (\mathbb{F}_p[X]/X^p)^{\times}$$
by sending $\tau$ to $1+X$, since $(1+X)^p = 1+X^p = 1 \in \mathbb{F}_p[X]/X^p$. But this representation $\rho$ does not lift to a map to $(\mathbb{F}_p[X]/X^{p+1})^{\times}$, since $\tau$ would have to map to $1+X+vX^p$, and
$$(1+X+vX^p)^p \equiv 1 + X^p \mod X^{p+1}$$
whereas $\tau^p$ should map to $1$. So there is an obstruction to lifting. Let us see this obstruction as a 2 cocycle of the adjoint representation, which is the trivial representation in dimension one. There is a set theoretic lift of $\rho$ given by
$$\rho(\tau^k) = (1+X)^k$$
for $k = 0,1,\ldots,p-1$. This leads to the $2$-cocycle
$$c(\tau^r,\tau^s) = \rho(\tau^r) \rho(\tau^s) \rho(\tau^{r+s})^{-1} = (1+X)^r (1+X)^s (1+X)^{-(r + s \bmod p)} = \begin{cases} 1, & r + s < p \\ (1+X)^p=1+X^p, & r+s \ge p \end{cases}$$ interpreting this as being viewed in $\mathbb{F}_p$ via the isomorphism $a \mapsto 1 + a X^p \bmod X^{p+1}$, this is exactly the "carry" $2$-cocycle which generates $H^2(G,\mathbb{F}_p)$.
As for Galois groups, you could take $G$ to be the absolute Galois group of $\mathbb{Q}_q$ for a prime $q$ with $p \| q-1$. Then the pro-p abelian quotient of $G$ is exactly $G = \mathbb{Z}/p \oplus \mathbb{Z}_p$, where the first factor is the inertia group generated by $\tau$. Then exactly as above the representation sending $\tau$ to $1 + X \mod (X^p,p)$ doesn't lift modulo $(X^{p+1},p)$ and the obstruction generates $H^2(G,\mathbb{F}_p) \simeq \mathbb{F}_p$.