A block of mass 0.25 kg is connected to a spring with spring constant 35 N/m. The block is oscillating on a frictionless horizontal surface. Its speed as it passes through its equilibrium position is 1.04 m/s. What's the total energy of the system?

To find the total energy of the system, we need to consider both kinetic energy and potential energy.

1. Kinetic Energy (KE): The kinetic energy is given by the formula KE = (1/2)mv², where m is the mass of the block and v is its velocity.

Given:
Mass of the block (m) = 0.25 kg
Velocity of the block (v) = 1.04 m/s

Using the formula, we can calculate the kinetic energy:
KE = (1/2)(0.25 kg)(1.04 m/s)²
KE = 0.1352 Joules

2. Potential Energy (PE): The potential energy is stored in the spring and is given by the formula PE = (1/2)kx², where k is the spring constant and x is the displacement of the block from its equilibrium position.

Given:
Spring constant (k) = 35 N/m

First, we need to calculate the displacement (x) of the block from its equilibrium position. Since the block is moving at a speed of 1.04 m/s as it passes through its equilibrium position, we know that the maximum displacement is half the amplitude of the oscillation (A), which can be calculated as A = vmax / ω, where vmax is the maximum velocity and ω is the angular frequency.

Given:
Maximum velocity (vmax) = 1.04 m/s

The angular frequency (ω) is given by ω = √(k/m), where k is the spring constant and m is the mass of the block.

Using the given values, we can calculate the angular frequency:
ω = √(35 N/m / 0.25 kg)
ω = √(140 N/kg)
ω ≈ 11.83 rad/s

Now, we can calculate the displacement:
A = vmax / ω
A = 1.04 m/s / 11.83 rad/s
A ≈ 0.0879 m

Finally, we can calculate the potential energy:
PE = (1/2)(35 N/m)(0.0879 m)²
PE ≈ 0.1342 Joules

The total energy of the system is the sum of the kinetic and potential energies:
Total Energy = KE + PE
Total Energy = 0.1352 Joules + 0.1342 Joules
Total Energy ≈ 0.2694 Joules

Therefore, the total energy of the system is approximately 0.2694 Joules.