Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at&nbs

A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at x = 5 cm 0.25 J. The spring constant of the spring is ________Nm^{-1}

Option: 1

50


Option: 2

___


Option: 3

__


Option: 4

__


Answers (1)

best_answer

Given

A=10cm

At any instant total energy for free oscillation remains constant=\frac{1}{2} \mathrm{kA}^2

\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \\ & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \quad \Rightarrow \mathrm{K}=\frac{0.25 \times 2}{\mathrm{~A}^2} \\ & \Rightarrow \mathrm{k}=\frac{0.50}{(10 \mathrm{~cm})^2}=\frac{0.50}{\left(10 \times 10^{-2}\right)}=\frac{0.50 \times 10^4}{100} \\ & \mathrm{k}=0.50 \times 100=50 \mathrm{~N} / \mathrm{m} \end{aligned}

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel
anam-khan

CSAB 2025 counselling dates out for special, supernumerary rounds through JEE Mains
anam-khan

'Why are they dying only in Kota': SC grills Rajasthan over surge in student suicides

Latest Question