When stretched by 25 cm , a spring stores 5 J of energy. It is...

The potential energy stored in the spring is given by $U = \dfrac{1}{2}kx^2$.

Substituting the given values, $U = 5 \text{ J} and$x = 25 \text{ cm} = 0.25 \text{ m}$:<br><br>$5 = \dfrac{1}{2} k (0.25)^2$<br><br>$5 = \dfrac{1}{2} k \left(\dfrac{1}{16}\right)

$k = 5 \times 32 = 160 \text{ N/m}$

The frequency of oscillation is $f = 5 \text{ Hz}$. The angular frequency is $\omega = 2\pi f = 10\pi \text{ rad/s}$.

Using the formula for angular frequency $\omega =$\sqrt{\dfrac{k}{m}}:

$\omega^2 =$\dfrac{k}{m}

$(10\pi)^2 =$\dfrac{160}{m}

$100\pi^2 =$\dfrac{160}{m}

Given $\pi^2 \approx 10$:

$1000 =$\dfrac{160}{m}

$m =$\dfrac{160}{1000} =$0.16 \text{ kg}$