Question

object of mass m=2.6 kg is dropped from rest at a height of h above a massless spring with spring constant k=494.3 N/m that is initially at its equilibrium position. The block falls onto the spring and compresses it a distance d=0.66 m below its equilibrium position before momentarily bringing the block to a stop. Find the height h from which the block was dropped (above the equilibrium point of the spring).

Answer 1

3.56 m

Step-by-step explanation:

As the object falls from height h, and then compresses the spring a distance of 0.66m, its initial potential energy is converted into elastics energy of the spring.

Let g = 9.81m/s2. Knowing that the total change in potential height is h + d, we have the following equation

`E_p = E_s`

`mg(h + d) = kd^2/2`

`2.6*9.81(h + 0.66) = 494.3*0.66^2/2`

`h + 0.66 = 4.22`

`h = 4.22 - 0.66 = 3.56 m`