What is wrong with this approach?
TL;DR You are forgetting that the kinetic energy is lost to the perfect inelastic collision. It can be shown from the conservation of momentum that in this particular example the total lost kinetic energy due to the collision is $K_\text{lost} = \frac{1}{2} M v^2$, where $M$ is mass of the added sand.
It does not matter whether the particle gained mass or velocity, the kinetic energy is always
$$K = \frac{1}{2} m v^2$$
The total work done must equal kinetic energy given to the added mass of sand plus kinetic energy lost due to the perfect inelastic collision
$$W = \frac{1}{2} M v^2 + K_\text{lost}$$
where $M$ is the total added mass of sand up until some point in time. You have already correctly identified that $K_\text{lost} = \frac{1}{2} M v^2$ from the conservation of energy. We can also show that from the conservation of momentum
$$(m \cdot v) + (dm \cdot 0) = (m + dm) v'$$
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 $F = v \cdot dm/dt$, $dm$ 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 lost 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 energy lost in inelastic collision is
$$\boxed{K_\text{lost} = \frac{1}{2} M v^2}$$
and this exactly equals what you have already observed.