If I lift a body with a force greater than its weight, what will happen to the excess energy provided to the body I will give an example to explain my question.
Case 1:

An elevator lifts body a with force equal to its weight for a distance $d$


*

*Energy given  to the  body (work  done)$=$ Weight $×$ $d$


*Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ Weight $× \ d$
Case 2:

An elevator lifts the same body with force equal to twice weight it’s for a distance $d$


*

*Energy given to the body (work done) $=$ $2 \ ×$ weight $×$ $d$


*Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ weight $×$ $d$
So doubled the amount of energy I gave to the body yet it’s capacity to do work by falling down has not changed.
Where is the excess energy the lift provided the body?
(I am in 11th grade so please make your explanation simple enough for me to understand.)
 A: In both cases, the work provided by the elevator is turned into kinetic energy of the object.
In the moment the elevator stops (and stops doing work), the object thus carries kinetic energy and therefor will continue flying upwards. Both of your potential-energy calculations are therefore wrong - the actual top height will be more than $d$.
And the object will naturally reach higher in the second case since it gains a larger amount of kinetic energy in the second case, causing more potential energy to be stored.

In regular elevators you might feel various lifting forces without "flying" upwards as the elevator stops. So you might feel that the answer I provided is incorrect. But remember to include the decceleration as well. The larger lifting force in certain elevators might be exerted for a shorter time as well, with deceleration beginning before the stop is reached. In the ideal scenario of the elevator keeping a constant speed and then suddenly stopping immediately, then you will experience the "flying".
A: Elevators go through a phase of acceleration, constant speed and then deceleration.
It is only during the acceleration phase that the force is greater than the weight of the elevator and contents. The 'excess' force is converted into kinetic energy.
During constant speed (by definition) the force is EQUAL to the weight (if it were greater then the acceleration would continue)
As you approach the destination level the force reduces to be less than the weight and hence the kinetic energy is reduced and is converted to the final PE required to reach the destination.
It is only if the deceleration rate is greater than 1 g (9.8m/s^2) that you would actually lift off the floor. Lifts always decelerate at a slower rate - so you only feel a slight 'lifting' of your weight on the floor.
You may also look at the meaning of 'jerk' in this context - which is a measure of the increase or decrease in acceleration, and gives you that feeling of a bump as the elevator starts or stops.
