# The graph $F - x$ is given, find the compression produced in the spring when a body of mass $5kg$ moving with velocity $8m/s$ hits the spring. Also calculate the force constant of the spring.

Answer

Verified

290.7k+ views

**Hint:**To find the force constant of the spring see the graph and find the slope. Then, calculate it by putting values. Now, to calculate the compression we have to use the conservation of energy in which total energy in an isolated system remains constant. So, kinetic and potential energy become equal to each other.

**Complete step by step answer:**

From the graph in the question, we can conclude that the slope of the graph is the force constant of the spring.

$

\because F = kx \\

\therefore k = \dfrac{x}{F} \\

$

where, $F$ is the force

$k$ is the force constant

$x$ is the compression of the spring

The slope of a graph can be calculated when we determine the difference between coordinates of the y – axis and x – axis respectively. After this, the differences of y – coordinates and x – coordinates are divided.

If we take $\left( {0.3m,24N} \right)$ and $\left( {0.2m,16N} \right)$ coordinates from x – axis and y – axis respectively, we get –

$

\implies Slope = \dfrac{{24 - 16}}{{0.3 - 0.2}} \\

\Rightarrow Slope = \dfrac{8}{{0.1}} \\

Slope = 80N/m \\

$

We know that, $slope = k$

Therefore, the force constant of the spring is $80N/m$.

From the question, we can conclude that the block will continue to compress till the block comes to the rest.

Now, using the conservation of energy the potential energy of the spring and kinetic energy of the spring becomes equal to each other because according to conservation of energy in an isolated system, the total energy remains constant and is said to be conserved over time.

$\therefore U = K \cdots \left( 1 \right)$

where, $U$ is the potential energy and $K$ is the kinetic energy

We know that, for a spring

$

\implies U = \dfrac{1}{2}k{x^2} \\

\implies K = \dfrac{1}{2}m{v^2} \\

$

From equation $\left( 1 \right)$, we get –

$\dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v^2}$

Cancelling $\dfrac{1}{2}$ on both sides, we get –

$

k{x^2} = m{v^2} \\

\implies x = v\sqrt {\dfrac{m}{k}} \cdots \left( 2 \right) \\

$

According to the question, it is given that –

Velocity, $v = 8m/s$

Mass, $m = 5kg$

Putting these values in equation $\left( 2 \right)$, we get –

$

\implies x = 8\sqrt {\dfrac{5}{{80}}} \\

\implies x = 8\sqrt {\dfrac{1}{{16}}} \\

\implies x = \dfrac{8}{4} = 2m \\

$

**Hence, compression produced by the spring after the hitting of the body is $2m$.**

**Note:**A spring stores potential energy due to extension. Since an unextended spring does not store potential energy, it is used as the point of zero energy. For a spring, potential energy is defined as, $U = \dfrac{1}{2}k{x^2}$ where, $x$ is the compression of the spring.

Last updated date: 03rd Jun 2023

•

Total views: 290.7k

•

Views today: 5.47k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main