DanDanDan
  • DanDanDan
A vertical spring with a force constant of 700 N/m is initially at equilibrium. A 10kg mass is attached to the end of the spring, causes it to elongate. What will be the maximum elongation/deformation of the spring? solve using work-energy theorem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
ma=1/2*kx^2 10*9.8=1/2*700*x^2 x approximately equals 0.529m
DanDanDan
  • DanDanDan
wer did u get this -> ma=1/2*kx^2
DanDanDan
  • DanDanDan
i mean the ma..

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anonymous
  • anonymous
ma=potential energy
anonymous
  • anonymous
F=ma
anonymous
  • anonymous
a is g
anonymous
  • anonymous
gravity
DanDanDan
  • DanDanDan
oww.. okok, yeah, i can remember it now,, gheez thanks !! ..
DanDanDan
  • DanDanDan
can i still ask one more??
anonymous
  • anonymous
kk
DanDanDan
  • DanDanDan
A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m. What is the coefficient of friction between the block and the tabletop?
anonymous
  • anonymous
(mu_k)mg(delta d)=1/2*kx^2
anonymous
  • anonymous
(mu_k)(0.5)(9.8)(1)=1/2*(100)(0.20)^2
anonymous
  • anonymous
(mu_k)=4/9.8
anonymous
  • anonymous
basically all the kinetic energy of the spring is converted into heat/friction
DanDanDan
  • DanDanDan
okok xD
DanDanDan
  • DanDanDan
wait. whats this --> (mu_k)
anonymous
  • anonymous
coefficient of friction
DanDanDan
  • DanDanDan
i think its just mu..
DanDanDan
  • DanDanDan
aww. okok, k stands for kinetic xDD

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