Final answer:
The ball's speed when it leaves the gun is 10.0 m/s.
Step-by-step explanation:
The speed at which the steel ball leaves the gun can be calculated using the principle of conservation of mechanical energy. The potential energy stored in the compressed spring is converted into kinetic energy of the ball when it is released.
The potential energy stored in the spring is given by the formula PE = (1/2)kx^2, where PE is the potential energy, k is the force constant of the spring, and x is the compression distance.
The kinetic energy of the ball when it leaves the gun is given by the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the ball, and v is its velocity.
Setting the potential energy equal to the kinetic energy and solving for v, we get:
v = sqrt((2kx^2)/m)
Plugging in the given values, we have:
v = sqrt((2 * 275 N/m * (6.9 cm)^2) / 0.01 kg) = 10.0 m/s
Therefore, the ball's speed when it leaves the gun is 10.0 m/s.