Solution
The platform moves away from the wall with constant speed $v$.
Therefore, the horizontal portion of the string between the pulley and
the wall increases in length at the rate $v$.
Hence, to keep the total string length constant, the inclined portion
must shorten at the same rate.
1. Velocity constraint
Let $u$ be the velocity of the block relative to the platform, towards the pulley. The rate at which the inclined segment shortens is the component of $u$ along the string:
$$u\cos\theta=v.$$
Hence,
$$
\boxed{u=v\sec\theta}.
$$
Although the platform has constant velocity, $u$ changes because
$\theta$ changes. Therefore, the block has a non-zero acceleration.
Differentiate:
$$
a=\frac{du}{dt}
=v\sec\theta\tan\theta\,\dot\theta.
$$
From the geometry, if the horizontal separation of the block from the
pulley is $x$,
$$
\tan\theta=\frac{h}{x}.
$$
Since the block approaches the pulley relative to the platform,
$$
-\frac{dx}{dt}=u.
$$
Differentiating $\tan\theta=h/x$,
$$
\sec^2\theta\,\dot\theta
=
\frac{hu}{x^2}.
$$
Using
$$
x=h\cot\theta,
$$
we obtain
$$
\boxed{\dot\theta=\frac{u\sin^2\theta}{h}}.
$$
Therefore,
\[
\begin{aligned}
a
&=v\sec\theta\tan\theta
\left(\frac{u\sin^2\theta}{h}\right)\\[4pt]
&=\frac{uv\sin\theta\tan^2\theta}{h}.
\end{aligned}
\]
Using $u=v\sec\theta$,
\[
a
=
\frac{v^2\sin\theta\tan^2\theta}
{h\cos\theta}.
\]
Since
$$
\frac{\sin\theta}{\cos\theta}\tan^2\theta
=\tan^3\theta,
$$
we get
$$
\boxed{a=\frac{v^2\tan^3\theta}{h}}.
$$
2. Equation of motion of the block
The block moves towards the right relative to the platform.
Therefore, friction on the block acts towards the left.
Let the tension in the string be $T$.
Vertically, the block has no acceleration, so
$$
N+T\sin\theta=mg,
$$
or
$$
N=mg-T\sin\theta.
$$
Hence the friction is
$$
f=\mu N
=\mu(mg-T\sin\theta).
$$
Taking rightward as positive for the block,
$$
T\cos\theta-f=ma.
$$
Therefore,
$$
T\cos\theta
-\mu(mg-T\sin\theta)
=ma.
$$
Thus,
\[
T(\cos\theta+\mu\sin\theta)
=
\mu mg+ma.
\]
Hence,
$$
\boxed{
T=
\frac{\mu mg+ma}
{\cos\theta+\mu\sin\theta}
}.
$$
3. Equation for the complete system
Now consider the block + platform together. The friction between the block and platform is internal and hence cancels. The resultant horizontal force exerted by the string on the complete system is $T$ towards the wall, while the external pulling force $F$ acts away from the wall. Since the platform has zero acceleration and only the block has horizontal acceleration $a$,
$$
T-F=ma.
$$
Therefore,
$$
F=T-ma.
$$
Substituting the value of $T$,
\[
F=
\frac{\mu mg+ma}
{\cos\theta+\mu\sin\theta}
-ma.
\]
Thus,
\[
F=
\frac{\mu mg}
{\cos\theta+\mu\sin\theta}
+
ma
\left[
\frac{1}
{\cos\theta+\mu\sin\theta}
-1
\right].
\]
Finally, using
$$
a=\frac{v^2\tan^3\theta}{h},
$$
we obtain
\[
\boxed{
F=
m\left[
\frac{
v^2\tan^3\theta+\mu gh
}
{
h(\cos\theta+\mu\sin\theta)
}
–
\frac{v^2\tan^3\theta}{h}
\right]
}
\]