Medium
When a $4\, kg$ mass is hung vertically on a light spring that obeys Hooke's law,the spring stretches by $2\, cm$. The work required to be done by an external agent in stretching this spring by $5\, cm$ will be ......... $J$ $(g = 9.8\, m/s^2)$.
- A$4.90$
- B$2.45$
- C$0.495$
- D$0.245$
Explore More
Similar Questions
$A$ block of mass $m$ initially at rest is dropped from a height $h$ onto a spring of force constant $k$. If the maximum compression in the spring is $x$,then:
Medium
View Solution$A$ spring of $5 \times 10^3 \text{ Nm}^{-1}$ spring constant is stretched initially by $10 \text{ cm}$ from its unstretched position. The work required to stretch it further by another $10 \text{ cm}$ is: (in $\text{ J}$)
$A$ spring of original length $L$ and spring constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Find the work done in the second stretching.
Difficult
View Solution$A$ body of mass $0.15 \ kg$ moving with a velocity of $15 \ ms^{-1}$ comes to rest when it hits a spring that is fixed at the other end. If the force constant of the spring is $1500 \ Nm^{-1}$,then the compression in the spring is: (in $m$)
$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is
MediumAIEEE 2005
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo