Answer:
The maximum velocity of the block is 2.56 m/s.
Explanation:
Given;
mass of the block, m₁ = 45g = 0.045 kg
spring constant, k = 240 N/m
displacement of the block, x = 3.5 cm = 0.035 m
Apply the principle of conservation of energy;
P.E = K.E
¹/₂kx² = ¹/₂m(v-u)²
where;
v is the maximum velocity of the block
u is the initial velocity of the block, u = 0
¹/₂kx² = ¹/₂mv²
kx² = mv²
[tex]v^2 = \frac{kx^2}{m}\\\\v = \sqrt{\frac{kx^2}{m}}\\\\v = \sqrt{\frac{(240) (0.035)^2 }{0.045}} \\\\v = 2.56 \ m/s[/tex]
Therefore, the maximum velocity of the block is 2.56 m/s.
