Question

A spring with a spring constant of 68 newtons per meter
hangs from a ceiling. When a 12-newton downward
force is applied to the free end of the spring, the spring
stretches a total distance of
(1) 5.7 m
(2) 0.59 m
(3) 820 m
(4) 0.18 m

Answer 1

Answer:

Option (4) 0.18 m

Explanation:

The following were obtained from the question:

K = 68N/m

F = 12N

e =?

F = Ke

e = F/k

e = 12/68

e = 0.18m

Answer 2

The spring  stretches a total distance of 0.18 m. And option (4) is correct.

Given data:

The value of spring constant is, k = 68 N/m.

The magnitude of downward force is, F = 12 N.

The spring constant is a numerical value that determines the stiffness of spring material. Such that the it is obtained by taking the ratio of spring force and distance of stretch. This is known as Hooke's law.

According to Hooke's law,

$F = k \times x\\\\x=\dfrac{F}{k} \\\\x=\dfrac{12}{68}\\\\x \approx 0.18 \;\rm m$

Thus, we can conclude that the spring  stretches a total distance of 0.18 m. And option (4) is correct.

Learn more about the spring force here:

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