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vfiekz [6]
3 years ago
7

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.
Physics
1 answer:
Eva8 [605]3 years ago
8 0

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

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