Height of the rise of the object attached to the spring ?

[Buffu]

Homework Statement

Question :- A block of mass $5 kg$ is attached to a spring. The spring is stretched by $10 cm$ under the load of the block. A impulse is provided to the block such that it moves up with a velocity of $2 m/s$. Find the height it will rise.

Homework Equations

$F_s = kx$
$x = 0.1m$
$v = 2m/s$
$m = 5kg$
$x_2$ - height raised from the mean position of the spring.

The Attempt at a Solution

We can get spring constant by equating spring force and weight of the object.
$$kx = mg $$
$$k = {5 \times 10 \over 1/10} = 500 \hspace{20 mm} (1)$$

Now initial total energy of the system :-

$$TE_i = {m v^2 + kx^2\over 2}$$
Now using given data and from $(1)$.
$$TE_i = {5 \times 4 + 500\times 1/100\over 2} = {25 \over 2}$$

final total energy :-

Velocity at maximum will be zero.

$$TE_f = {mg(x + x_2) + kx_2^2\over 2} \hspace{20 mm} (2)$$

Using conservation of energy ,$(1)$ and $(2)$ we get,

$${25 \over 2} = {mg(x + x_2) + kx_2^2\over 2}$$
$$20x_2^2 + 2x_2 -0.8 = 0$$

$x_2 = 0.15$
thus total height is $0.25 m$.
The answer is $0.20 m$. I think the answer is incorrect, since this is pretty straight forward question.

 

[Doc Al]

In Equation (2) you divided the gravitational PE by 2.

 

[Buffu]

In Equation (2) you divided the gravitational PE by 2

Such a silly mistake and i am trying this question for more than an hour.

 

[Doc Al]

Such a silly mistake and i am trying this question for more than an hour.

That's the story of my life!