Solved

The Potential Energy of a Spring with a Constant Of k

Question 28

Short Answer

The potential energy of a spring with a constant of kk , a maximum amplitude of AA , and an angular velocity of ω\omega is given by P=12kA2cos2ωtP=\frac{1}{2} k A^{2} \cos^{2} \omega t
(joules) where tt is in seconds. Find the rate of change in the potential energy of a mass on a spring with a constant of 1.2 N/m1.2 \mathrm{~N} / \mathrm{m} , a maximum amplitude of 0.53 m0.53 \mathrm{~m} , and an angular velocity of 0.031rad/s0.031 \mathrm{rad} / \mathrm{s} , when t=5.8 st=5.8 \mathrm{~s} .

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions

Unlock this Answer For Free Now!

View this answer and more for free by performing one of the following actions

qr-code

Scan the QR code to install the App and get 2 free unlocks

upload documents

Unlock quizzes for free by uploading documents