This question is somewhat vague, as it does not specify the work done by whom (or what.) Let's therefore consider the total work done on the system. The total work done on the system equals the change in the energy of the system:
`W = E_f - E_i` . Here, `E_i` and `E_f` are initial and final total energy of the system, respectively. This is known as work-energy theorem.
Let's now consider the energy change in the following processes.
1. A 10 kg weight rests on the table.
Since the weight is at rest, neither its potential energy nor its kinetic energy is changing. The work done is W = 0.
2. A person holds a 1 kg weight 1 m away from the floor. Again, here the weight is at rest, so its potential energy and kinetic energy do not change. The total work done is W = 0.
3. A person lifts a 1 kg weight to 1 m away from the floor. Here, the gravitational potential energy of the weight changes. The change equals `DeltaU = mgh = 1 kg * 9.8 m/s^2 * 1 m = 9.8 J` . So, the work done on the weight is W = 9.8 J.
4. A 10 kg ball is rolled across the floor at a constant speed for a distance of 10 m.
Assuming the floor is horizontal, the gravitational potential energy of the ball does not change during this process. Since the speed remains constant, the kinetic energy `K = (mv^2)/2` does not change as well. The work done on the ball is again W = 0.
(This might seem counterintuitive - if the ball is rolling, should not someone perform the work to make it move, and keep it at a constant speed? However, if someone does apply a force with a horizontal component to the ball, it would get accelerated according to the first Newton's Law `vecF = mveca` . Then, the ball's speed would change. Since the speed is constant, it means there is another force acting on the ball, such as friction. The total work of the external force and the friction force is zero.)
Thus, the process 3 requires the most work, W = 9.8 J.
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