91影视

According to this law, the total energy of the system remains constant.

${\mathbf{K}}_{\mathbf{1}}\mathbf{+}{\mathbf{U}}_{\mathbf{1}}\mathbf{=}{\mathbf{K}}_{\mathbf{2}}\mathbf{+}{\mathbf{U}}_{\mathbf{2}}$

Here,${\mathit{K}}_{\mathbf{1}}$ is the initial kinetic energy,${\mathit{U}}_{\mathbf{1}}$ is the initial potential energy,${\mathit{K}}_{\mathbf{2}}$ is the final kinetic energy,${\mathit{U}}_{\mathbf{2}}$ is the final potential energy.

Let the speed of at the bottom be v, for the object of mass m. From the conservation of energy, the potential energy at the top will be equal to the kinetic energy at the bottom.

$$\begin{gathered} mgh=\frac{1}{2}m{v}^2 \\ {v}^2=2gh \\ v=\sqrt{2gh} \end{gathered}$$

From the above equation, it can be observed that the speed of the object is independent of the mass of the falling body.

Therefore, all the masses will have the same speed at the bottom.

From the work-energy theorem, the work done is equal to the change in kinetic energy. Therefore, the work done will be maximum when the kinetic energy is maximum at the bottom. It is known that the velocity is same for all the masses, so the kinetic energy will be maximum for the heavier object.

Therefore, the maximum work will be done for the object of mass 2m.