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# A block of mass 5.0kg is suspended from the end of a vertical spring which is stretched by 10cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed so 2.0m/s. How high will it rise? (Take g=10$m/{s^2}$).

Last updated date: 09th Aug 2024
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Hint: Here we will be using the law of conservation of energy. According to which, the energy can neither be created nor be destroyed, it can change from one form to another. We know that the spring force is directly proportional to the displacement from the initial position.

Formula used:
${F_s} = - kx$
Where, k is equal to the force constant
${F_s}$=spring force and x is equal to the displacement of spring from the initial position.

Complete step by step solution:
We are provided with the following data:
Mass of the block, $m=5kg$
A vertical spring is stretched by $10cm$ that is $x=10cm=0.1m$.
Acceleration due to gravity, $g=10m/{s^2}$.
Then we are interested to find the height $h$.
We know that the spring force is directly proportional to the displacement x from the equilibrium position.
That is given by, ${F_s}\alpha x$
${F_s} = - kx$
Where $k$ is called the spring constant.
From this equation, we can calculate the spring constant for the given spring in this problem.
We know that before applying the force block is at rest.
So on rearranging the equation we get, $k = \dfrac{F}{x} = \dfrac{{mg}}{x}$
$\Rightarrow k = \dfrac{{5 \times 10}}{{10}} = 500N{m^{ - 1}}$
We know that energy is conserved. Thus the total energy (from initial position) just after the force is applied is represented as $E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}k{x^2}$ ……………….. (1)
However, when the impulsive force is applied the block is given a velocity 2m/s in the upward direction.
So in other words, at that moment, the total energy is the sum of potential energy and kinetic energy due to the impulsive force.
Total energy at a height $h = \dfrac{1}{2}k{(h - x)^2} + mgh$ …………….. (2)
We have two equations which are equal to energy. To solve further equate the LHS of equation (1) and (2) we get,
$\Rightarrow \dfrac{1}{2}m{v^2} + \dfrac{1}{2}k{x^2} = \dfrac{1}{2}k{(h - x)^2} + mgh$
On further simplification we get,
$\Rightarrow \dfrac{1}{2}m{v^2} + \dfrac{1}{2}k{x^2} = \dfrac{1}{2}k{h^2} - khx + \dfrac{1}{2}k{x^2} + mgh$
Let's solve it further and simplify the equation, we get,
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{2}k{h^2}$
Rearrange the equation concerning the height h, we get
$\Rightarrow {h^2} = \dfrac{{m{v^2}}}{k}$
$\Rightarrow {h^2} = \dfrac{{5 \times {2^2}}}{{500}} = \dfrac{{20}}{{500}}$
Take the square root of the equation we get the height h, we get
$\Rightarrow h = \sqrt {\dfrac{1}{{25}}} = 0.2m$
Thus $h= 20cm$.

$\therefore$ 20m is the height of the stretched string when suspended from the load.

(i) $k$ value is large if the spring is stiff and $k$ value is small if the spring is soft.