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

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A spring with a spring constant of 68 newtons per meter hangs from a ceiling. Whena 12-newton downward force is applied to the free end of the spring, the spring streches a total distance of

(1) 5.7m

(2) 0.59m

(3) 820m

(4) 0.18m

Answer: (4) 0.18 m

Explanation: The force necessary to compress or extend a spring by a displacement is directly proportional to that distance and the spring's constant. This relationship is stated by a principle called Hooke's Law:

[tex]F=-k\Delta x[/tex]

k is spring constant, which shows how stiff the spring is and so, is characteristic of each spring;

x is total displacement;

The negative sign on the formula indicates the restoring force caused by the spring is in opposite direction related to the force and it is why the displacement occurs.

So, for the spring in the question above:

[tex]F=-k\Delta x[/tex]

[tex]\Delta x=\frac{F}{k}[/tex]

[tex]\Delta x=\frac{12}{68}[/tex]

[tex]\Delta x = 0.18[/tex]

When a 12 N force is applied to the spring, it streches 0.18 m.

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