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
299.7k+ 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.
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.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

