Q24P - Fundamentals Of Physics

Question

A block of mass $m\text{ = 2.0}$kg is dropped from height $\text{h=40 cm}$ onto a spring of spring constant $\text{k=1960 N/m}$ (Figure). Find the maximum distance the spring is compressed.

Step-by-Step Solution

Answer

Maximum spring compression is $0.10\text{m}$ .

Step 1: Given

1. Mass of blocks, $m=20\mathrm{kg}$ 
2. Height from which the block is dropped,  $h=4\text{0cm}=0.40\text{m}$
3. Spring constant,  $K=1960\text{N/m}$

Step 2: Determining the Concept

The problem is based on the law of conservation of energy, which states that the total energy of an isolated system remains constant.Using the given situation and the conservation of energy, find the spring compression.According to the law of energy conservation, energy can neither be created nor be destroyed.

 

Formula:

1. Potential energy,  $PE=mgh$
2. Elastic energy,  $E=\frac{1}{2}{K}^2$
3. $PE+KE=\text{constant}$
4. Root of a quadratic equation,  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where, KE is kinetic energy, PE is potential energy, m is mass, v is velocity, g is an acceleration due to gravity, x is displacement, K is spring constant, and W is work done.

Step 3: Determining the maximum spring compression

From the situation given in the problem, at the top point, the block has potential energy, and in the compressed state, it has elastic energy.

So, from the law of conservation of energy,

 $$\begin{gathered} {\left(PE+KE\right)}_{top}={\left(PE+KE\right)}_{compressed} \\ mg\left(h+x\right)+0=0+\frac{1}{2}k{x}^2 \\ 0=-\left[mg\left(h+x\right)\right]+\frac{1}{2}k{x}^2 \\ -mgh-mgx+\frac{1}{2}k{x}^2=0 \\ \left(k\right)x^2+\left(-2mg\right)x+\left(-2mgh\right) \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{2mg\pm \sqrt{4m^2{g}^2+8kmgh}}{2k} \end{gathered}$$

 

 Plugging all the values properly,

 $x=0.099\text{9m}$

  $x=0.10m$

 

Hence, maximum spring compression is $\text{0.10m.}$