Question
Download Solution PDFThe stiffness of a spring is 20 N/cm. Calculate the work done to stretch the spring by 7 cm.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The work done by the spring is calculated by:
\(W=\frac {1}{2} KX^2\)
where W = work done in Joules
K = stiffness in N/cm
X = displacement in cm
Calculation:
Given:
K = 20 N/cm
X = 7 cm
Work done = \(\frac {1}{2} \times 20 \times7^2 = 490\)N cm = 4.9 joule
So work done is 4.9 joule.
Last updated on May 9, 2025
-> PGCIL Diploma Trainee result 2025 will be released in the third week of May.
-> The PGCIL Diploma Trainee Answer key 2025 has been released on 12th April. Candidates can raise objection from 12 April to 14 April 2025.
-> The PGCIL DT Exam was conducted on 11 April 2025.
-> Candidates had applied online from 21st October 2024 to 19th November 2024.
-> A total of 666 vacancies have been released.
-> Candidates between 18 -27 years of age, with a diploma in the concerned stream are eligible.
-> Attempt PGCIL Diploma Trainee Previous Year Papers for good preparation.