A 0.455 kg wood block is firmly attached to a very light horizontal spring (k = 160 N/m). It is noted that the block-spring system, when compressed 5.0 cm and released, stretches out 2.3 cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table

Answer 1

If there is no friction, the block will travel to 5 cm beyond the equilibrium point before turning back.

k=160 N/m
mass, m = 0.455 kg
coefficient of kinetic friction = μ
F=kinetic frictional force
=μm

Potential energy of spring when compressed 5 cm
E1 = (k/2)*(0.05)²
Potential energy of spring when stretched 2.3 cm
E2 = (k/2)*(0.023)²
Lost potential energy
=E1-E2
The energy is lost to overcoming the frictional force over a distance of
D = (5.0 + 2.3) cm
= 0.073 m
Work done by friction
=μ*m*D

Equate energies,
μ*m*D = E1-E2
Solve for μ

Answer 2

The solution was wrong.
E1 is the only energy to consider for the spring, not E2.
E1 = μ*m*D was enough for solving this problem.

Good luck!

Answer 3

E1=mu*mg*D

Answer 4

you need to figure this out yourself