Question

A 0.5k block attached to a spring (k=200N/m) is sitting on top of a rough horizontal surface. Initially a spring is stretched 10 cm from its equillibrium position. When released, the block moves and stops 8 cm past the equillibrium position in the other direction. Find the friction coefficient between the block and surface.

Answer 1

Initial energy of the block = (1/2)Kx²
where K is spring constant = 200 N/m
and x is the initial amount of stretching = 10 cm = 0.1 m
= (1/2)*200*(0.1)² = 1 J ---------------(1)
Final energy of the block = Potential energy of spring + Friction energy
= (1/2)KX² + W_f
Where X is compression of the spring = 8 cm = 0.08 m
W_f is the energy lost due to friction
Final energy = (1/2)*200*(0.08)² +W_f = 0.64 + W_f -------------(2)
applying energy conservation
Initial energy = Final energy
0.64 + W_f =1  
W_f = 0.36 J
Now, We know that energy lost due to friction is given by
W_f = Friction Force*distance = (u*R_N)*d
from the free body diagram
R_N = mg = 0.5*9.81 = 4.905 N and d = 18 cm = 0.18 m
W_f = u*(4.905)*0.18 = 0.883*u
u = W_f/0.883 = 0.36/0.883 = 0.41