Question

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

Answer 1

ma=1/2*kx^2
10*9.8=1/2*700*x^2
x approximately equals 0.529m

Answer 2

wer did u get this -> ma=1/2*kx^2

Answer 3

i mean the ma..

Answer 4

ma=potential energy

Answer 5

F=ma

Answer 6

a is g

Answer 7

gravity

Answer 8

oww.. okok, yeah, i can remember it now,, gheez thanks !! ..

Answer 9

can i still ask one more??

Answer 10

kk

Answer 11

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?

Answer 12

(mu_k)mg(delta d)=1/2*kx^2

Answer 13

(mu_k)(0.5)(9.8)(1)=1/2*(100)(0.20)^2

Answer 14

(mu_k)=4/9.8

Answer 15

basically all the kinetic energy of the spring is converted into heat/friction

Answer 16

okok xD

Answer 17

wait. whats this --> (mu_k)

Answer 18

coefficient of friction

Answer 19

i think its just mu..

Answer 20

aww. okok, k stands for kinetic xDD