Answer:
Maximum speed of the block, [tex]v_{max} = 4.58 m/s[/tex]
Explanation:
Mass of the block, m = 2.0 kg
Spring constant, k = 600 N/m
Spring extension, x = 20 cm = 0.2 m
Speed of the block due to the extension, v = 3.0 m/s
First, Potential energy, PE stored in the spring:
PE = 0.5 kx²
PE = 0.5 * 600 * 0.2²
PE = 12 J
Calculate the kinetic energy of the block due to the extension:
[tex]KE_x = 0.5 mv^2\\KE_x = 0.5 * 2 * 3^2\\KE_x = 9 J[/tex]
The maximum Kinetic Energy of the block will be:
[tex]KE_{max} = 0.5 m v_{max}^2\\KE_{max} = 0.5 * 2 * v_{max}^2\\KE_{max} = v_{max}^2[/tex]
[tex]KE_{max} = KE_x + PE\\v_{max}^2 = 9 + 12\\ v_{max}^2 = 21\\ v_{max} = \sqrt{21} \\ v_{max} = 4.58 m/s[/tex]
