A vertical spring (ignore its mass), whose spring constant is 594-N/m, is attached to a table and is compressed down by 0.196-m. What upward speed (in m/s) can it give to a 0.477-kg ball when released?

Respuesta :

Answer:

Speed, v = 6.91 m/s

Explanation:

Given that,

Spring constant, k = 594 N/m

It is attached to a table and is compressed down by 0.196 m, x = 0.196 m

We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,

[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]

[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]

v = 6.91 m/s

So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.