Answer:
The maximum compression of the spring after the collision is 0.15 m
Explanation:
Given data
Mass of the block (m) = 0.80 kg
Initial velocity (v) = 1.2 m/s
Spring constant (k) = 50 N/m
Find the maximum compression of the spring (x) after compression
Potential energy of the spring = Kinetic energy of the block
Kinetic energy of the block = 0.5 × (mv)²
Kinetic energy of the block = 0.5 × (0.80 × 1.2)²
Kinetic energy of the block =0.5 × 0.9216
Kinetic energy of the block = 0.4608 ---------->(1)
Potential energy of the spring = 0.5 × k × x²
Potential energy of the spring = 0.5 × 50 × x²
Potential energy of the spring = 25 x² ---------> (2)
Equate (1) and (2)
25 x² = 0.4608
x² = 0.018432 m²
x =0.1357 = 0.15 m
Therefore the maximum compression of the spring after collision is 0.15 m
Critical angle=
Arcsin(sqrt(eps2/eps1))
Where eps1 is relative permittivity of the fiber and eps2 is for the medium just outside the fiber (coating)
Note for this angle to exist
eps1>eps2 since the sin function is bound between-1 and 1. Thus the fiber is more optically dense than the jacket.
Answer: This isn't a question but the statement is quite accurate to a well thought out science lesson.