A lift with the total mass of 2200 kg was elevated by an electric motor from the initial level at 4m to 80m above the floor of the shaft. Steel cables holding the lift broke when the lift reached the 80m level. A linear amortisation spring of uncompressed ("relaxed") length equal to 4m is installed at the floor of the shaft. Neglect friction. Calculate:
(a) the work done by the electric motor to elevate the lift from the level at 4m to the level at 80m,
(b) the velocity of the free falling lift at the moment of contact with the spring (at the level of 4m),
(c) the minimum stiffness of the spring (see the last part of question (d)),
(d) the total energy of the lift–spring system
• at the initial state of the lift at the level of 4m,
• at the 80m level of the lift,
• at the moment of contact of the lift with the spring during the fall,
• at the lowest lift level corresponding to the maximum compression of the spring, which is 1m above the shaft floor.

a)  W = 1,639 10⁶ J, b)     v = 38.60 m / s, c)    k = 3.78 10⁵ N / m

Explanation:

a) In this part they ask us the work done by the motor, let's use the work energy theorem

            W = ΔEm = Em_f - Em₀

            W = mh y₂ - mgy₁

            W = mg (y₂ -y₁)

            W = 2200 9.8 (80 -4)

            W = 1,639 10⁶ J

b) for this part let's use conservation of energy

starting point. Highest point

            Em₀ = U = m g y₂

final point . Punoit in contact with the spring

           Em_f = K + U

           Em_f = ½ m v² + m g y₁

           Em₀ = Em_f

           m g y₂ = ½ m v² + mg y₁

           v = √(2g (y₂ - y₁))

           v = √(2 9.8 (80-4))

           v = 38.60 m / s

c) the stiffness of the spring

starting point. Just when it comes into contact with the spring

           Em₀ = K + U

           Em₀ = ½ m v² + mgy₁

Final point. With the spring at maximum compression x = 3 m

           Em_f = Ke + U

           Em_f = ½ k x² + m g y₃

           Em₀ = Em_f

           ½ m v² + mgy₁ = ½ k x² + m g y₃

           ½ k x² = ½ m v² + m g (y₁ - y₃)

let's calculate

            ½ 3² k = ½ 2200 38.60² + 2200 9.8 (4 -1)

            4.5 k = 1.639 10⁶ +6.468 10⁴ = 1.7036 10⁶

            k = 1.7036 10⁶ / 4.5

             k = 3.78 10⁵ N / m

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

where’s the list ? the question is incomplete