Question

Block A of mass m hangs from a spring of spring constant k in equilibrium. At equilibrium block A is given vertically downward velocity v so that maximum extension in spring is 2 mg/k Value k of vo, is
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Answer 1

Answer:

An imaginary plane shut placed at focus perpendicular to the principal axis is termed as focal plane. It is considered to be existed but not actually.

Answer 2

Value of k is $\frac{4mg^2}{v_0^2}$

Explanation:

Given:

A mass m hangs with a spring in equilibrium

Block is given a vertical velocity $v_0$ so that the maximum extension in the spring is 2mg/k

To find out:

The value of k

Solution:

When the spring is stretched to the maximum, the potential energy stored in the spring will be equal to the change in kinetic energy of the block

Initial kinetic energy of the block = 0

We know that for a spring with spring constant k and stretched by x, the potential energy is given by

$U=\frac{1}{2}kx^2$

$\implies \frac{1}{2}k\times (\frac{2mg}{k})^2=\frac{1}{2}mv_0^2-0$

$\implies \frac{4mg^2}{k}=v_0^2$

$\implies k=\frac{4mg^2}{v_0^2}$

Hope this answer is helpful.

Know More:

Q: A block of mass m is suspended through a spring of  spring constant k and is in equilibrium. A sharp blow  gives the block an initial downward velocity v. How far  below the equilibrium position, the block comes to an  instantaneous rest?

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