Question

A 0.300 kg mass is connected to a horizontal spring, with simple friction between the mass and the surface having a coefficient of µk = 0.0800. The force constant of the spring is k = 50.0 N/m. What distance would the mass travel if it were released 0.150 m from equilibrium with zero initial velocity?
__________m

Answer 1

stored energy in spring = 0.5*k*x^2
                                          = 0.5*50*0.15^2
                                          = 0.5625 J
let the distance travelled by x'm
Final energy stored = 0.5*k*(0.15-x')^2
energy lost by friction = miu*m*g*x'

use:
initial energy = final energy + energy lost in friction
0.5625 = 0.5*k*(0.15-x')^2 + miu*m*g*x'
0.5625 = 0.5*50*(0.15-x')^2 + 0.08*0.3*9.8*x'
0.5625 = 0.5*50*(0.0225 + x'^2 - 0.3x') + 0.08*0.3*9.8*x'
0.5625 = 0.5625+ 25*x'^2 - 7.5 x' + 0.2352*x'
25x'^2 - 7.2648x' = 0
25x'- 7.2648= 0
x'= 0.29 m
Answer: 0.29 m