Question

A 1.1-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.

Answer 1

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 ,$g=9.8m/s^2$

$x=\frac{F}{k}=\frac{1.1\cdot 9.8}{160}=0.0674 m$

$x=0.0674\times 100=6.74 cm (1m=100 cm)$

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.

$K.E=\frac{1}{2}kx^2$ (Law of conservation of energy)

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Substitute the values then we get

$1.1\cdot v^2=160\cdot (0.26)^2$

$v=\sqrt{\frac{160\cdot(0.26)^2}{1.1}}$

$v=3.14 m/s$

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