Answer :
Answer:
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 ,[tex]g=9.8m/s^2[/tex]
[tex]x=\frac{F}{k}=\frac{1.1\cdot 9.8}{160}=0.0674 m[/tex]
[tex]x=0.0674\times 100=6.74 cm (1m=100 cm)[/tex]
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.
[tex]K.E=\frac{1}{2}kx^2[/tex] (Law of conservation of energy)
[tex]\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex]
Substitute the values then we get
[tex]1.1\cdot v^2=160\cdot (0.26)^2[/tex]
[tex]v=\sqrt{\frac{160\cdot(0.26)^2}{1.1}}[/tex]
[tex]v=3.14 m/s[/tex]
Hence, the speed of an object with which the object passes through its original position on the way up=3.14 m/s