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kondor19780726 [428]

1 year ago

6

49. A vertically hung 0.50-meter- long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight a

ttached to the spring. If 15 joules of elastic potential energy are stored in the spring , what is the value of the spring constant?

1 answer:

Natali5045456 [20]1 year ago

4 0

Answer:

$K=120$

Explanation:

From the question we are told that

Length of spring $l_1=0.5m$

Length of stretched $l_s=1m$

Potential energy of spring $E=15J$

Generally equation for energy stored is mathematically given as

$U=1/2K \triangle x^2$

$K=\frac{2U}{\triangle x^2}$

$K=\frac{2*15}{ 0.5^2}$

Therefore value of the spring constant in N/m? is given as

$K=120$

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