A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
The blocks shown are released from rest with spring unstretched. Pulley & horizontal surface are frictionless. If k=400 N/m and M=4.5 kg, what is maximum extension of spring?
http://s3.amazonaws.com/answerboardimage/20071222120396333222723984175004973.jpg
anonymous
 3 years ago
The blocks shown are released from rest with spring unstretched. Pulley & horizontal surface are frictionless. If k=400 N/m and M=4.5 kg, what is maximum extension of spring? http://s3.amazonaws.com/answerboardimage/20071222120396333222723984175004973.jpg

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where are you stuck? during max extension, they are not moving. for M, \(\Sigma F_y= TF_g=0\) \(T=F_g=4.5*9.8\) for 2M, \(\Sigma F_x= TF_s=0\) \(F_s=T=4.5*9.8\) \(kx=4.5*9.8\) sub k and solve for x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi! Careful: this assumption is not correct: "during max extension, they are not moving. for M, \(\Sigma F_y= TF_g=0\)" Velocity is zero, but they are accelerating, so \(\Sigma F_y\) is not zero. Best way to solve this problem is to use conservation of mechanical energy.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah. missed that. sorry. i was thinking that it was not SHM

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the conservation of ME, since there is no potential energy, or kinetic energy, and setting that height of M as 0 at the very beginning, and that at the max extension, the velocities are zero, i.e. still no kinetic energy, \(0=\frac{1}{2} kx^2 + MgX\) sub it all in and this should do it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.