Question

A block of mass, 10 kg, collides with a spring with a spring constant of 200 N/m at a speed of 12 m/s compressing the spring. When the block stops moving, how far is the spring compressed?​

Answer 1

AnswEr :

• String is compressed at 2.68 m far.

GivEn :

• Mass of the block = 10 kg

• Spring Constant = 200 N/m

• Speed = 12 m/s

To find :

• How far does the spring gets compressed when the block stops moving = ?

SoluTion :

According to the law of conservation of energy,

$\dashrightarrow \: \: \: {\sf{\purple{ KE_{block} = PE_{spring}}}}$

$\dashrightarrow \: \: \:\sf \purple{\dfrac{1}{2}m_{block} {v}^{2}_{block} = \dfrac{1}{2} k {( \Delta x)}^{2} }$

• $\sf m_{block}$ is the mass of the block

• $\sf v_{block}$ is the speed of the block

• k is constant

• ∆x is the distance

So, total distance it covered :

$: \implies \: \: \sf \: \Delta x = \sqrt{ \dfrac{m_{block} {v}^{2}m_{block} }{k}}$

Substituting the values,

$: \implies \: \: \sf \: \Delta x = \sqrt{ \dfrac{ 10 \times {12}^{2} }{200}}$

$: \implies \: \: \sf \: \Delta x = \sqrt{ \dfrac{ 10 \times 144 }{200}}$

$: \implies \: \: \sf \: \Delta x = \sqrt{ \dfrac{ 1440 }{200}}$

$: \implies \: \: \sf \: \Delta x = \sqrt{ 72}$

$: \implies \: \: \sf \: { \underline{ \boxed{ \sf{ \red{\Delta x = 2.68 \: m \: (approx).}}}}} \: \bigstar$

Additional information :

• Law of Conservation of Energy :

This law states that, energy can neither be created nor be destroyed. Although, it may be transformed from one form to another .

Example :

Let us take the example of above question, KE of the block before it collides with the spring is converted into PE of the spring before it gets compressed . In this case the KE is transformed into PE.