The diagram on figure below shows a cart on the ground pulling the 20 kg out of a 4 m/s instant. A uniform speed along a horizontal surface, using a stride that

Mass $m = 100 \, \text{g} = 0.1 \, \text{kg}$.
Distance (height) $h = 100 \, \text{m}$.
Time $t = 10 \, \text{s}$.
Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$.
Since the body is moving uniformly, the applied force $F$ is equal to its weight.

$$F = mg$$ $$F = (0.1 \, \text{kg})(9.8 \, \text{m/s}^2)$$ $$F = 0.98 \, \text{N}$$

Work done $W$ is the force multiplied by the distance moved in the direction of the force.

$$W = Fh$$ $$W = (0.98 \, \text{N})(100 \, \text{m})$$ $$W = 98 \, \text{J}$$

The work done is $\boxed{\text{98 J}}$.

Power $P$ is the work done divided by the time taken.

$$P = \frac{W}{t}$$ $$P = \frac{98 \, \text{J}}{10 \, \text{s}}$$ $$P = 9.8 \, \text{W}$$

The power is $\boxed{\text{9.8 W}}$.

The force required to hoist the load is the output force, and the distance is the height.

$$W_{out} = F_{load} \times h$$ $$W_{out} = 500 \, \text{N} \times 4 \, \text{m}$$ $$W_{out} = 2000 \, \text{J}$$

The efficiency of the hoist gear ($\eta_{gear}$) is the ratio of the work output from the gear to the energy input to the gear.

$$\eta_{gear} = \frac{W_{out}}{E_{input\_gear}}$$ $$E_{input\_gear} = \frac{W_{out}}{\eta_{gear}}$$ $$E_{input\_gear} = \frac{2000 \, \text{J}}{0.80}$$ $$E_{input\_gear} = 2500 \, \text{J}$$

The energy input to the hoist gear is the energy output from the motor ($E_{output\_motor}$).
The efficiency of the motor ($\eta_{motor}$) is the ratio of the energy output from the motor to the energy input to the motor.

$$\eta_{motor} = \frac{E_{output\_motor}}{E_{input\_motor}}$$ $$E_{input\_motor} = \frac{E_{output\_motor}}{\eta_{motor}}$$ $$E_{input\_motor} = \frac{2500 \, \text{J}}{0.70}$$ $$E_{input\_motor} \approx 3571.43 \, \text{J}$$

The input energy to the hoist motor is $\boxed{\text{3571.43 J}}$.