Answer:
(a) the spring constant = 107.3N/m
(b) the mass (in kg) of the block = 0.0531kg
(c) the frequency (in Hz) of oscillation =
7.15Hz
Step-by-step explanation:
An oscillating block-spring system has a mechanical energy of 0.569 J, an amplitude of 10.3 cm, and a maximum speed of 4.63 m/s.
(a) the spring constant
The formula for finding the spring constant k
k = 2E / x²
Where E = mechanical energy = 0.569J
x = amplitude = 10.3cm
We convert 10.3cm to meter(m)
100cm = 1m
10.3cm = ?
= 10.3cm ÷ 100cm
= 0.103m
Spring constant (k) = (2 × 0.569j) ÷ (0.103)²
Spring constant (k) = 107.3N/m
(b) the mass (in kg) of the block
Using the Kinectic Energy = 1/2m(V²max)
Therefore, the formula for Mass(m) = 2E/(V²max)
V = Maximum speed = 4.63 m/s
Mass(m) = (2× 0.569J) ÷ (4.63)²
Mass(m) = 0.0531kg
(c) the frequency (in Hz) of oscillation
The formula of Frequency of oscillation = F = (1/2π) × √k/m
Where k = spring constant = 107.3N/m
m = mass of the block = 0.0531kg
F = (1 ÷ 2π) × √ ( 107.3N/m ÷ 0.0531kg)
F = 7.15Hz