When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
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Answer:
B = 62.9 N
Explanation:
This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid
B = ρ g V
write the equilibrium equation
T + B -W = 0
B = W- T (1)
use the density to write the weight
ρ = m / V
m = ρ V
W = ρ g V
substitute in 1
B = m g -T
B = g V - T
To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³
calculate
B = 7850 9.8 1.20 10⁻³ - 29.4
B = 92.3 - 29.4
B = 62.9 N