Heavy duty springs are always installed on the ground at the bottom of the shaft underneath an elevator to provide last-minute rescue. (if you have a chance, take a look at our school elevator, you'll see them located on the ground at b level). in a "worst-case" design scenario, a 1550 kg elevator with broken cables is falling at 8.60 m/s just before it starts compressing a cushioning spring at the bottom. the spring is supposed to stop the elevator, compressing 26.0 cm as it does so.

find the spring constant. ignore air resistance and friction in the elevator guides.

k = 1.7 10⁶ N / m

Explanation:

Let's use the concepts of conservation of mechanical energy for this problem. Let's look for energy at two points, one before touching the springs and the other with compressed springs

Initial

      Em₀ = K = ½ m v²

Final

      = Ke = ½ k x²

As there is no friction the mechanical energy is conserved

     Em₀ =  

     ½ m v² = ½ k x²

     k= m v² / x²

     k = 1550 8.60² /0.260²

     k = 1.7 10⁶ N / m

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

ill do it for 100, also this isnt 80 its 40