A1.1-kg object is suspended from a vertical spring whose spring constant is 160 n/m. (a) find the amount by which the spring is stretched from its unstrained length. (b) the object is then pulled straight down by an additional distance of 0.26 m and released from rest. find the speed with which the object passes through its original position on the way up.

3.14 m/s

Explanation:

We are given that

Mass of object=1.1 Kg

Spring constant=160 N/m

a.We have to find the amount by which the spring is stretched from its unstrained length(it means x)

We know that F=kx ,

b.When an object is pulled straight down by an additional distance 0.26 m and released from rest.

x=0.26 m

We have to find the speed with which the object passes through its original position on the way up.

(Law of conservation of energy)

Substitute the values then we get

Hence, the speed of an object with which the object passes through its original position on the way up=3.14 m/s