Excellent!
Now, before we continue, I'm going to give you a little more information about reversible and irreversible processes, so that you can better understand what is happening to the gas in each case.
REVERSIBLE PROCESSES
As we said earlier, in a reversible process, the system experiences a deformational and thermal path that is only slightly removed from a continuous sequence of thermodynamic equilibrium states. For a thermodynamic equilibrium state of a gas within a cylinder, the temperature, pressure, and specific volume (reciprocal of density) of the gas are uniform throughout the cylinder, and the compressive stresses (forces per unit area) within gas are the same in all directions; for an ideal gas:
$$\sigma_x=P=\frac{RT}{V}$$
$$\sigma_y=P=\frac{RT}{V}$$
$$\sigma_z=P=\frac{RT}{V}$$where the ##\sigma##'s are the compressive stresses in the coordinate directions, T is the uniform temperature of the gas within the cylinder, p is the uniform pressure, and V is the total volume of the gas in the cylinder divided by the number of moles (i.e., the uniform specific volume)
IRREVERSIBLE PROCESSES
The mechanical behavior of a gas, in general, is more complicated than the (equilibrium) equation of state of the gas; it depends not only on the P-V-T behavior of the gas, but also on the rate at which the gas is being deformed. The latter is described by the spatial derivatives of the velocity of the gas. The more general equations for the compressive stresses in a gas that is not in thermodynamic equilibrium (derived in 3D from Newton's Law of viscosity) are given by:
$$\sigma_x=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_x}{\partial x}$$
$$\sigma_y=P-2\mu\frac{\partial u_y}{\partial y}=\frac{RT}{V}-2\mu\frac{\partial u_y}{\partial y}$$
$$\sigma_z=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z}$$where the u's are the velocity components in the coordinate directions and ##\mu## is the viscosity of the gas. In these equations, we still call P the pressure of the gas, although this "thermodynamic pressure" is no longer the total force per unit area acting on every surface; the local pressure P just contributes to the force per unit area. Note also that, in an irreversible process, the temperature, pressure, and specific volume of the gas in these equations are no longer uniform spatially within the cylinder (e.g., the specific volume is no longer the total volume of the cylinder divided by the total number of moles); all three are functions of the spatial coordinates x, y, and z. Thus, T = T(x,y,z), P = P(x,y,z), and V = V(x,y,z). Note also that, when the velocity components of a gas become very small, as in a reversible deformation, the equations for the stresses reduce to those for a reversible process.
There are several reasons why the temperature, pressure, and specific volume are not uniform in an irreversible process. First of all, the gas has mass, so it has to be accelerated just like the piston is accelerated. But it is not solid like the piston. So, in a rapid deformation, it experiences a kind of "sloshing" effect, where different parts of the gas experience different velocities and accelerations. Secondly, heat transfer is occurring at a finite rate from the reservoir to the gas, so portions of the gas near the wall experience temperature changes at a different rate than portions of the gas closer to the axis. This also causes the viscosity of the gas (which is temperature dependent) to affect the compressive stresses in the gas non-uniformly. Finally, there can be small scale turbulence in the gas as it deforms, and this influences the effective value of the viscosity (which feeds back into the compressive stresses).
The bottom line is that, in an irreversible process, many complicated phenomena are occurring within the gas that affects its behavior in a difficult-to-quantify manner, and we can't simply say that the compressive stress acting on the piston face can be determined by merely using the ideal gas equation under the assumption that the pressure, temperature, and specific volume are uniform throughout.
Now, back to our problem. The equation you wrote for the force exerted by the gas on the piston face was as follows:
$$F(t) = m(g + a(t))$$where we now know that ##F(t)=\sigma_zA=(\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z})A##, with the spatial derivative of vertical gas velocity evaluated at the piston face. Another way of writing your equation is:
$$F(t)=m(g+\frac{du}{dt})$$where u is the piston velocity. If we multiply this equation on both sides by u=dz/dt, we obtain:
$$F(t)\frac{dz}{dt}=m(g\frac{dz}{dt}+u\frac{du}{dt})$$
If I integrate the left hand side of this equation with respect to t, from time zero to time t, I get:
$$W(t)=\int_0^t{F(t')\frac{dz'}{dt'}dt'}=\int_{z_0}^{z(t)}{F(z')dz'}$$where W(t) is the work done by the gas on the piston from time zero (at which the pile of rocks was suddenly removed) to time t, z' and t' are dummy variables of integration, and ##z_0## is the elevation of the piston at time zero. What do you get when you integrate the right hand side of the equation with respect to t, and then combine the results?