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

Sagot :

okpalawalter8

Answer:

[tex]K=120[/tex]

Explanation:

From the question we are told that

Length of spring   [tex]l_1=0.5m[/tex]

Length of stretched [tex]l_s=1m[/tex]

Potential energy of spring [tex]E=15J[/tex]

Generally equation for energy stored is mathematically given as

[tex]U=1/2K \triangle x^2[/tex]

[tex]K=\frac{2U}{\triangle x^2}[/tex]

[tex]K=\frac{2*15}{ 0.5^2}[/tex]

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

[tex]K=120[/tex]

onyebuchinnaji

The value of the spring constant is 120 N/m.

Spring constant

The spring constant of the spring is calculated by applying the principle of conservation of energy as follows;

[tex]\frac{1}{2} k \Delta x = U[/tex]

Where;

  • K is the spring constant
  • Δx is the extension
  • U is the elastic energy

The value of the spring constant is calculated as follows;

[tex]k = \frac{2U}{\Delta x} \\\\k = \frac{2\times 15}{0.5^2}\\\\k = 120 \ N/m[/tex]

Learn more about spring constant here: https://brainly.com/question/1968517

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