A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.15 102 N/m that lies on a horizontal frictionless surface as shown in the figure below. The block is pulled to a position xi = 5.75 cm to the right of equilibrium and released from rest.

(a) Find the the work required to stretch the spring.
______ J

(b) Find the speed the block has as it passes through equilibrium.
_______ m/s

To find the work required to stretch the spring, we can use the formula for the potential energy stored in a spring:

Potential energy = (1/2) k (change in length)^2

In this case, the change in length is the displacement of the block from its equilibrium position, which is 5.75 cm or 0.0575 m. Plugging in the values, we get:

Potential energy = (1/2) * (5.15 * 10^2 N/m) * (0.0575 m)^2

Simplifying the equation, we get:

Potential energy = (1/2) * (515) * (0.00330625)

Therefore, the potential energy is approximately 0.540 J.

Now, to find the speed of the block as it passes through equilibrium, we can use the principle of conservation of mechanical energy. When the block is at its maximum displacement, all its potential energy is converted into kinetic energy.

Potential energy = Kinetic energy

Therefore:

(1/2) k (change in length)^2 = (1/2) m v^2

Plugging in the values, we get:

(1/2) * (5.15 * 10^2 N/m) * (0.0575 m)^2 = (1/2) * (2.00 kg) * v^2

Simplifying the equation, we get:

(1/2) * (515) * (0.00330625) = (1/2) * (2.00) * v^2

Solving for v, we get:

v^2 = [(515 * 0.00330625) / 2.00]

Taking the square root of both sides, we get:

v ≈ √(1.6928875)

Therefore, the speed of the block as it passes through equilibrium is approximately 1.30 m/s.