- #1
Ike
- 8
- 1
A 0.164-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 62 N/m. The block sticks to the spring and the spring compresses 0.17 m before coming to a momentary halt. What is the speed of the block just before it hits the spring? Give your answer in (m/s).
My initial thought was to use a conservation of energy formula. I came up with:
This, however, is not the correct answer. Therefore, there must be another way of doing this problem. Can anyone help me?
My initial thought was to use a conservation of energy formula. I came up with:
(1/2)(k)(x)^2 = (1/2)(m)(v)^2
(1/2)(62 N/m)(0.17 m)^2 = (1/2)(0.164 kg)(v)^2
v = {[(62 N/m)(0.17 m)^2]/(0.164 kg)}^(1/2)
(1/2)(62 N/m)(0.17 m)^2 = (1/2)(0.164 kg)(v)^2
v = {[(62 N/m)(0.17 m)^2]/(0.164 kg)}^(1/2)
This, however, is not the correct answer. Therefore, there must be another way of doing this problem. Can anyone help me?