Question

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?

Answer 1

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

$K=120$