Question

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?

Short answer

1. Work done by the block on the spring is $7.22J$.
2. Work done by the spring on theblock is$-7.22J$.
3. Value of $h_0$is 0.86m.
4. Maximum compression of the spring when the block is released from a height $2.0h_0$is 0.26m.

Step-by-step solution

Step 1: Given

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

Determining the concept

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.

(a) Determining the work done by the block on the spring

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 $7.22J$.

(b) Determining the work done by the spring on the block

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 $-7.22J$.

(c) Determining the value of h0

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$

(d) Determining the maximum compression of spring when the block is released from height 2.0h0

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.