Calculating Energy of the Landing Gear Retraction

The energy required is independent of the time. Find the centre of mass for the main strut + wheel assembly. Find the mass of the main strut + wheel assembly, call this M. Calculate the difference in height of the centre of mass when lowered and when raised, call this h.

The energy required is then PE= M x g x h, where g is the acceleration due to gravity (about 9.8 ms^-2).

As the 30 second time restraint means motion will be quite slow, kinetic energy will be negligible. Also, unless the side struts are made of some remarkably dense material, their difference in potential energy between lowered and raised will also be negligible.

The power required depends on the time, and is equal to PE ÷ t, where t is the time.

Whoever set this question does not really understand aircraft undercarriage design. The length of thee main strut equates to a plane the size of an executive jet or an island hopper. There are two main connection nuts (6) as stated but one is directly behind the one shown in the drawing so cannot be seen which is important as the two nuts have completely different functions and completely different designs, the drive (5) is normally hydraulic because an electric motor could not withstand to loading of an off center landing directly and the size of gear box needed would be too heavy to install, linkage (2&3) would buckle under landing impact and regulations require that this item be duplicated, there appears to be no shock absorber so the wing would fall off on the first landing. Not only does the question setter not understand the topic, he does not even seem to know how to set questions. The force required depends not just on the weight which isn't given, but also on the distribution of the weight which also isn't given.