A0.71-kg block attached to a spring with force constant112N/m is free to move on
ID: 2992787 • Letter: A
Question
A0.71-kg block attached to a spring with force constant112N/m is free to move on a frictionless, horizontal surface as in the figure below. The block is released from rest after the spring is stretched 0.13 m.
(a) At that instant, find the force on the block.
Explanation / Answer
a) Force = kx = 112*0.13 = 14.56 N to the left ma = kx a = 14.56/0.71 = 20.5 m/s^2 b) Force = kx = 112*0 = 0 a = 0 c) Force = kx = 112*0.13 = 14.56 N to the right ma = kx a = 14.56/0.71 = 20.5 m/s^2
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.