A 1.225-newton weight stretches a spring 0.6125 m. The weight is released from rest 1.5 m above the equilibrium position

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899604
Joined: Mon Aug 02, 2021 8:13 am

A 1.225-newton weight stretches a spring 0.6125 m. The weight is released from rest 1.5 m above the equilibrium position

Post by answerhappygod »

A 1 225 Newton Weight Stretches A Spring 0 6125 M The Weight Is Released From Rest 1 5 M Above The Equilibrium Position 1
A 1 225 Newton Weight Stretches A Spring 0 6125 M The Weight Is Released From Rest 1 5 M Above The Equilibrium Position 1 (26.59 KiB) Viewed 39 times
Writing -5/2 instead of -3/2 is also labelled as wrong.
A 1.225-newton weight stretches a spring 0.6125 m. The weight is released from rest 1.5 m above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to t times the instantaneous velocity. Use the Laplace transform to find the equation of motion x(t). (Use g = 9.8 m/s2 for the acceleration due to gravity.) 3 -(1) ✓15 X(t) = -() :-) COS ✓15 21 215 e sin 2 2 2 X eBook
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Top
Post Reply

Return to “Archived Questions & Answers”