What is wrong with this approach?
TL;DR You are correct that the force needed to maintain constant velocity of the cart is
$$F = \frac{dp}{dt} = \frac{d}{dt} m v = m \frac{dv}{dt} + v \frac{dm}{dt} = v \frac{dm}{dt} \qquad \text{where} \qquad \frac{dv}{dt} = 0 \tag 1$$
which follows directly from the second Newton's law of motion, but you are forgetting that the kinetic energy is lost to the perfect inelastic collision. I show below how to calculate the lost kinetic energy in two ways: (i) via conservation of energy, and (ii) via conservation of momentum.
Lost kinetic energy via conservation of energy
The work $dW$ done by the force $F$ over displacement $dx$ is
$$dW = F \cdot dx = v \frac{dm}{dt} dx = v \frac{dx}{dt} dm = v^2 \cdot dm$$
Integrating the above equation we get that the total work done by the force $F$ to give sand of mass $M$ required velocity $v$ is
$$W = M v^2$$
It does not matter whether the particle gained mass or velocity, the kinetic energy is always $K = \frac{1}{2} m v^2$. The change in kinetic energy in your example is
$$\Delta K = \frac{1}{2} (m + M) v^2 - \frac{1}{2} m v^2 = \frac{1}{2} M v^2$$
where $m$ is mass of empty cart. From the work-energy theorem and law of conservation of energy, kinetic energy lost to the perfect inelastic collision is
$$\boxed{K_\text{lost} = W - \Delta K = \frac{1}{2} M v^2}$$
Lost kinetic energy via conservation of momentum
Conservation of momentum for perfect inelastic collision is
$$(m \cdot v) + (dm \cdot 0) = (m + dm) v'$$
where $m$ is mass of the cart plus sand added up until some point in time, $v$ is the cart velocity maintained at constant value by applying force from Eq. (1), $dm$ is additional infinitesimally small mass of sand we put into the cart, and $v'$ is new velocity due to the perfect inelastic collision
$$v' = \frac{m}{m + dm} v$$
The infinitesimally small loss of kinetic energy due to the added mass of sand $dm$ is
$$dK = \frac{1}{2} m v^2 - \frac{1}{2} (m + dm) v'^2 = \frac{1}{2} m v^2 \frac{dm}{m + dm}$$
By rearranging the above equation we get
$$m \cdot dK + \underbrace{dm \cdot dK}_{\approx 0} = \frac{1}{2} m v^2 \cdot dm$$
We can neglect the product of two infinitesimally small values ($dm \cdot dK \approx 0$) which gives
$$dK = \frac{1}{2} v^2 \cdot dm$$
By integrating the above equation we finally get that the total kinetic energy lost in the perfect inelastic collision is
$$\boxed{K_\text{lost} = \frac{1}{2} M v^2}$$