The spring of a spring gun is compressed a distance d of 3.2 cm from its relaxed state, and a ball of mass m = 12g is put in the barrel. With what speed will the ball leave the barrel once the gun is fired? The spring constant is 7.5 N/cm. Assume no friction and a horizontal gun barrel.

the work done compressing the spring becomes the kinetic energy of the projectile

1/2 k x^2 = 1/2 m v^2

in cgs system
... 750 kdynes (3.2)^2 = 12 v^2 ... cm/s

To determine the speed at which the ball will leave the barrel, we can use the principle of conservation of mechanical energy.

First, we need to find the potential energy stored in the spring when it is compressed. The formula for potential energy stored in a spring is given by:

Potential energy (P.E.) = (1/2)kx^2

Where k is the spring constant and x is the displacement of the spring from its relaxed state.

In this case, the spring constant is 7.5 N/cm, which is equivalent to 0.075 N/mm (since 1 cm = 10 mm). The displacement of the spring, x, is 3.2 cm, which is equivalent to 32 mm.

P.E. = (1/2) * 0.075 N/mm * (32 mm)^2
P.E. = 0.075 * 512 N * mm
P.E. = 38.4 N * mm

Next, we can consider the conversion of potential energy into kinetic energy. The formula for kinetic energy is given by:

Kinetic energy (K.E.) = (1/2)mv^2

Where m is the mass of the ball and v is its velocity.

In this case, the mass of the ball, m, is 12 g, which is equivalent to 0.012 kg.

We can equate the potential energy stored in the spring to the kinetic energy of the ball:

38.4 N * mm = (1/2) * 0.012 kg * v^2

Next, we can convert the units of potential energy from N * mm to joules (J) by dividing by 1000 (since 1 N * m = 1 joule).

38.4 N * mm = 0.0384 N * m = 0.0384 J

Substituting this value into the equation:

0.0384 J = (1/2) * 0.012 kg * v^2

Now, we can solve for v:

v^2 = (2 * 0.0384 J) / 0.012 kg
v^2 = 3.2 m^2/s^2

Taking the square root of both sides:

v = √(3.2 m^2/s^2)
v = 1.79 m/s

Therefore, the ball will leave the barrel with a speed of approximately 1.79 m/s.