Step 1: Use the general variable-mass equation instead of the bare force-equals-rate-of-momentum shortcut.
For a body whose mass is changing, Newton's second law generalises to
\[ F_{ext} + u\frac{dm}{dt} = m\frac{dv}{dt} \]
where $u$ is the velocity of the incoming (or outgoing) mass relative to the main body, and $dm/dt$ is the rate at which mass joins it.
Step 2: Identify each term for this problem.
The truck's own velocity does not change, since uniform velocity is given, so $dv/dt = 0$ and the equation reduces to $F_{ext} = -u\,dm/dt$. The sand falls in vertically, so before it lands its velocity in the truck's direction of motion is zero, meaning its velocity relative to the truck is $u = 0 - v = -v$.
Step 3: Convert the mass rate to SI units.
\[ \frac{dm}{dt} = 24 \ \text{kg min}^{-1} = \frac{24}{60} = 0.4 \ \text{kg s}^{-1} \]
Step 4: Substitute into the equation.
\[ F_{ext} = -u\frac{dm}{dt} = -(-v)(0.4) = v \times 0.4 \]
With $v = 15$ m/s:
\[ F_{ext} = 15 \times 0.4 = 6 \ \text{N} \]
Step 5: Physical meaning.
This external force is exactly what is needed to speed up each new chunk of sand from rest, relative to the ground in the horizontal direction, up to the truck's 15 m/s, every second, so the truck itself never slows down.
Final Answer:
\[ \boxed{6 \ \text{N}} \]