A small steel ball bearing with a mass of 10 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.19 m. Calculate the speed at which the ball leaves the spring.

potential energy= Kinetic energy at release

mgh=1/2 mv^2

v=sqrt(2gh)

To calculate the speed at which the ball leaves the spring, we can use the principle of conservation of mechanical energy. The mechanical energy of the system is conserved, meaning the initial mechanical energy is equal to the final mechanical energy.

The initial mechanical energy consists of two parts: the potential energy due to the compression of the spring and the kinetic energy of the ball bearing. The potential energy can be calculated using Hooke's Law, which states that the potential energy stored in a spring is given by the equation:

PE = (1/2)kx^2

where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.

The kinetic energy of the ball bearing can be given by the equation:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the ball bearing, and v is its velocity.

Since the ball bearing is released from rest, its initial kinetic energy is zero.

When the ball bearing reaches its maximum height, all of its kinetic energy is converted to potential energy. Therefore, at the maximum height, its final kinetic energy is zero and its potential energy is equal to the initial total mechanical energy.

The total mechanical energy can be expressed as:

E = PE + KE

Since the initial kinetic energy is zero, E = PE.

Now let's calculate the potential energy at the maximum height:

PE = mgh

where m is the mass of the ball bearing, g is the acceleration due to gravity, and h is the maximum height.

Substituting the values into the equation:

PE = (0.01 kg)(9.8 m/s^2)(1.19 m)
= 0.11742 J

Now, equating the potential energy to the total mechanical energy:

0.11742 J = (1/2)kx^2

To find the spring constant, k, we need to know more information about the spring. Assuming it is a simple harmonic spring, we can estimate its spring constant.

Let's say the spring constant, k, is 100 N/m. Now we can solve for x, the displacement of the spring:

0.11742 J = (1/2)(100 N/m)x^2

Simplifying the equation:

x^2 = (0.11742 J) / (0.5 * 100 N/m)
= 0.0023484 m^2

Taking the square root of both sides:

x = sqrt(0.0023484 m^2)
= 0.04844 m

Now we have the displacement of the spring, x.

Finally, we can calculate the speed at which the ball leaves the spring using the equation for potential energy:

PE = (1/2)kx^2

Since the kinetic energy is zero at the maximum height, we can rearrange the equation to solve for the final velocity:

v = sqrt((2/m) * PE)

Substituting the values:

v = sqrt((2/0.01 kg)(0.11742 J))
= sqrt(23.484 m^2/s^2)
= 4.846 m/s

Therefore, the speed at which the ball leaves the spring is approximately 4.846 m/s.