Question

A 2.0-kg block sliding on a frictionless, horizontal surface is attached to one end of a horizontal spring (k = 600 N/m) which has its other end fixed. The speed of the block when the spring is extended 20 cm is equal to 3.0 m/s. What is the maximum speed of this block as it oscillates?

Answer 1

Answer:

Maximum speed of the block, $v_{max} = 4.58 m/s$

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:

$KE_x = 0.5 mv^2\\KE_x = 0.5 * 2 * 3^2\\KE_x = 9 J$

The maximum Kinetic Energy of the block will be:

$KE_{max} = 0.5 m v_{max}^2\\KE_{max} = 0.5 * 2 * v_{max}^2\\KE_{max} = v_{max}^2$

$KE_{max} = KE_x + PE\\v_{max}^2 = 9 + 12\\ v_{max}^2 = 21\\ v_{max} = \sqrt{21} \\ v_{max} = 4.58 m/s$