A block of mass m= 7.00kg is attached to the end of an ideal spring. Due to the
ID: 1294104 • Letter: A
Question
A block of mass m= 7.00kg is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h= 9.00cm from its equilibrium length. The spring has an unknown spring constant k. Take the acceleration due to gravity to be g = 9.81m/s2 .
What is the spring constant k?
Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency ? of the block's oscillations about its equilibrium position.
Please show how you got the answer :) Thank you!
Explanation / Answer
F = k*x...
here F= m*g = 7*9.81 =68.67 N..
x = 9 cm =0.09 m....
spring constant k = F/x = 68.67/0.09 = 763 N/m......
w = sqrt(k/m) = sqrt(763/7) = 10.44 rad/sec
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.