A mass and spring are arranged on a horizontal, frictionless table as shown in the figure below. The spring constant is k = 480 N/m, and the mass is 4.8 kg. The block is pushed against the spring so that the spring is compressed an amount 0.31 m, and then it is released. Find the velocity of the mass when it leaves the spring.

k=(mv2)/h2, where h is the compressed amount

so v=square roof of (k*(h)squared)/m

square root(480*0.0961)/4.8=3.1 m/s

To find the velocity of the mass when it leaves the spring, we can use the concept of conservation of mechanical energy.

1. The potential energy of the compressed spring can be calculated using the formula: PE = (1/2)kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
Substituting the given values: PE = (1/2)(480 N/m)(0.31 m)^2

2. The potential energy converts into kinetic energy as the mass is released from the spring. Therefore, the potential energy when the mass leaves the spring is equal to the kinetic energy. Using the formula: KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Substituting the given values: PE = KE = (1/2)(4.8 kg)v^2

3. Setting the potential energy equal to the kinetic energy: (1/2)(480 N/m)(0.31 m)^2 = (1/2)(4.8 kg)v^2

4. Solving the equation for v^2: (480 N/m)(0.31 m)^2 = (4.8 kg)v^2

5. Simplifying the equation: (0.4656 N)(0.4656 m) = v^2

6. Taking the square root of both sides to find v: v = √(0.4656 N)(0.4656 m)

7. Calculating the value of v using a calculator: v ≈ 0.682 m/s

Therefore, the velocity of the mass when it leaves the spring is approximately 0.682 m/s.

To find the velocity of the mass when it leaves the spring, you can use the principle of conservation of mechanical energy.

First, let's find the potential energy of the spring when it is compressed by 0.31 m. The potential energy stored in a spring is given by the equation:

Elastic potential energy (PE) = (1/2)kx^2

where k is the spring constant and x is the displacement from the equilibrium position.

Given: k = 480 N/m and x = 0.31 m

Plugging in these values, we have:

PE = (1/2) * 480 N/m * (0.31 m)^2

Next, let's equate the potential energy stored in the spring to the kinetic energy of the mass when it leaves the spring. The kinetic energy of an object is given by the equation:

Kinetic energy (KE) = (1/2)mv^2

where m is the mass of the object and v is its velocity.

Since mechanical energy is conserved, we can write:

PE = KE

Plugging in the values we know:

(1/2) * 480 N/m * (0.31 m)^2 = (1/2) * 4.8 kg * v^2

Now, we can solve for v by rearranging the equation:

v^2 = (480 N/m * (0.31 m)^2) / 4.8 kg

v^2 = (72.048 N * m) / 4.8 kg

v^2 = 15.01 N * (m/kg)

v = √(15.01 N * (m/kg))

Calculating this, we find that v ≈ 3.88 m/s.

Therefore, the velocity of the mass when it leaves the spring is approximately 3.88 m/s.