Answer:
b = 0.6487 kg / s
Explanation:
In the context of oscillatory motion, friction is related to velocity,
fr = - b v
where b represents the friction coefficient.
Upon solving the equation, the angular velocity is represented as
w² = k / m - (b / 2m)²
In this case, we're given an angular frequency w = 1Hz, the mass m = 0.1 kg, and the spring constant k = 5 N / m. This allows us to derive the friction coefficient.
Let’s denote
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Now, let's calculate the angular frequencies.
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
Substituting values yields
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.
