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

The maximum speed of the block as it oscillates can be calculated using the concept of conservation of mechanical energy. The potential energy stored in the spring is converted into kinetic energy when the block is released. At the maximum amplitude, all of the potential energy is transformed into kinetic energy, resulting in the maximum speed of 3.46 m/s.

Step-by-step explanation:

The maximum speed of the block as it oscillates can be calculated using the concept of conservation of mechanical energy. When the block is released from rest, the potential energy stored in the spring is converted into kinetic energy. At the maximum amplitude, all of the potential energy is transformed into kinetic energy, resulting in the maximum speed.

The potential energy stored in the spring can be calculated using the formula:

PE = (1/2) k x2

Where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the spring is extended by 20 cm, which is equal to 0.2 m. The spring constant is given as 600 N/m.

Plugging in the values:

PE = (1/2) * 600 * (0.2)2 = 12 J

Since the potential energy at the maximum amplitude is equal to the kinetic energy, we can calculate the maximum speed using the formula:

KE = (1/2) mv2

Where KE is the kinetic energy, m is the mass of the block, and v is the maximum speed.

Plugging in the values:

12 J = (1/2) * 2 * v2

v2 = 12 J / 1 kg = 12 m2/s2

v = √(12 m2/s2) = 3.46 m/s

Therefore, the maximum speed of the block as it oscillates is 3.46 m/s.

Answer 2

Maximum speed of the block,
`v_(max) = 4.58 m/s`

Step-by-step 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) = √(21) \\ v_(max) = 4.58 m/s`