A 700gm block is released from rest at height _{${\mathit{h}}_{\mathbf{0}}$} above a vertical spring with spring constant $\mathit{k}\mathbf{=}\mathbf{400}\mathit{N}\mathbf{/}\mathit{m}$and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 19.00 cm . How much work is done?

1. By the block on the spring
2. By the spring on the block
3. What is the value of ${\mathit{h}}_{\mathbf{0}}$?
4. If the block were released from height $\mathbf{2}\mathbf{.}\mathbf{00}{\mathit{h}}_{\mathbf{0}}$above the spring, what would be the maximum compression of the spring?

1. Mass of block,$m=0.7Kg$
2. Spring constant,$K=400N/m$
3. Compression of spring,$x_j=0.19m$

First, find the work done by the block and the spring by using the elastic potential formula. Using the law of conservation of energy, find the initial height of the block. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formula are as follow:

1. Work done,$W=PE=-\frac{1}{2}k{x}^2$
2. Total energy,$E_{total}=KE+PE$
3. $KE=\frac{1}{2}m{v}^2$
   
4. $PE=mgh$
   

where, KE is kinetic energy, PEis potential energy, W is work done, m is mass, v is velocity, g is an acceleration due to gravity, is total energy, x is displacement, k is spring constant, and h is height.

Work done by the block on thespring is given by elastic potential energy,

$\begin{array}{rcl}W& =& \frac{1}{2}k{x_j}^2\\ W& =& \frac{1}{2}\left(400\right){\left(0.19\right)}^2\\ W_b& =& 7.22J\end{array}$

Hence, work done by the block on the spring is $7.22J$.

The work done by the spring on theblock is in the opposite direction of thework done by the block on thespring.

So, the answer is,

$W_s=-7.22J$
Hence, work done by the spring on the block is $-7.22J$.

Since the total energy is conserved, therefore,

$\begin{array}{rcl}& & \left(\text{changein}PE\right)\text{+}\left(\text{changein}GPE\right)\text{=0}\\ \left(\frac{1}{2}k{\left(x_f-x_i\right)}^2+mg\left(h_f-h_i\right)\right)& =& 0\\ \left(\frac{1}{2}k\left(x_f^2\right)-mg{h}_i\right)& =& 0\\ \frac{1}{2}k\left(x_f^2\right)& =& mg{h}_i\\ h_i& =& \frac{\frac{1}{2}k{x}_f^2}{mg}\\ h_0& =& 1.052\end{array}$

Now, the above value of height is from the starting point to the compressed point of the spring which is at 0.19m below. So, the original height is

$h_i=h_0-x$

$$\begin{gathered} h_i=1.052-0.19 \\ {h}_i=0.86m \end{gathered}$$

Hence the value of$h_0$is$0.86m$$h_0$

As, given in the problem,

$2h_0=2.104m$

So, at top point,thepotential energy and during compression elastic potential occurs.

So,

$\begin{array}{rcl}\frac{1}{2}k{x}^2& =& mg{h}_0\\ x^2& =& \frac{2mg{h}_0}{k}\\ x^2& =& \frac{2\left(0.7\right)\left(9.8\right)\left(2.104\right)}{400}\\ x& =& 0.26m\end{array}$

Hence, maximum compression of the spring when the block is released from a height $2.0h_0$is 0.26m.