Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A spring of spring constant  5\times 10^{3} N/m is stretched initially by  5 cm from the unstretched position. Then the work required to stretch it further by another  5 cm is

Answers (1)

\\\text{work done in stretching a spring by a distance x is } \\ W=\frac{1}{2} k x^{2}$ \\ where $\mathrm{k}$ is the spring constant.

\\\text{work required to stretch the string by 10 cm is }\\ W_{1}=\frac{1}{2} k x^{2}=\frac{1}{2} 5 \times 10^{3}(0.1)^{2}=25 J$ \\ work required to stretch the string by $5 \mathrm{cm}$ is

\\W_{2}=\frac{1}{2} k x^{2}=\frac{1}{2} 5 \times 10^{3}(0.05)^{2}=6.25 J$ \\ Hence the work done in stretching it from $5 \mathrm{cm}$ to $10 \mathrm{cm}$ is \\\\ $W_{1}-W_{2}=25-6.25=18.75 J$

Posted by

lovekush

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question