A horizontal spring attached to a wall has a force constant of 780 N/m. A block of mass 1.70 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. a)Find the energy stored in the spring when the mass is stretched 6.20 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. b)Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. c)What is the speed at the halfway point?

To find the velocity at the equilibrium point after the block is released, we need to determine the energy stored in the spring when the mass is stretched 6.20 cm from equilibrium and find the speed of the mass as it passes through equilibrium.

a) To find the energy stored in the spring when the mass is stretched 6.20 cm from equilibrium, we can use the formula for potential energy stored in a spring:

Potential Energy (PE) = 0.5 * k * x^2

Where k is the force constant of the spring and x is the displacement from equilibrium.

PE = 0.5 * 780 N/m * (0.062 m)^2
= 0.1215 J

So the potential energy stored in the spring when the mass is stretched 6.20 cm from equilibrium is 0.1215 J.

Now, let's find the energy stored in the spring when the mass passes through equilibrium after being released from rest. At the equilibrium point, the mass has maximum kinetic energy and zero potential energy.

Thus, the energy stored in the spring at the equilibrium point is equal to the initial potential energy.

PE = 0.1215 J

b) To find the speed of the mass as it passes through equilibrium, we can use the conservation of energy equation:

Total Energy (TE) = Potential Energy (PE) + Kinetic Energy (KE)

At the equilibrium point, the total energy is equal to the sum of the potential energy and the kinetic energy:

TE = PE + KE

Since the potential energy at the equilibrium point is the same as the initial potential energy, the equation becomes:

TE = 2 * PE

TE = 2 * 0.1215 J
= 0.243 J

Since the total energy at the equilibrium point is the sum of the kinetic energy and the potential energy, we can write:

0.243 J = KE + 0.1215 J

Solving for KE:

KE = 0.243 J - 0.1215 J
= 0.1215 J

The kinetic energy at the equilibrium point is 0.1215 J.

Now, we can use the equation for kinetic energy:

KE = 0.5 * m * v^2

Where m is the mass of the block and v is the velocity of the block.

0.1215 J = 0.5 * 1.70 kg * v^2

Simplifying and solving for v:

v^2 = (0.1215 J * 2) / (0.5 * 1.70 kg)
v^2 = 0.143 J / kg
v^2 = 0.143 m^2/s^2

Taking the square root of both sides:

v = sqrt(0.143 m^2/s^2)
v = 0.379 m/s

So the velocity of the mass at the equilibrium point is 0.379 m/s.

c) To find the speed at the halfway point, we can use the concept of conservation of mechanical energy. At the halfway point of the oscillation, the block has equal amounts of kinetic and potential energy.

Thus, the speed at the halfway point is equal to the velocity at the equilibrium point, which is 0.379 m/s.