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 = 12 g 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.
force=k*3.2=7.5*3.2 N
then
a= forceinN/massinkg
To find the speed at which the ball leaves the barrel, we can use the principle of conservation of mechanical energy.
First, let's determine the potential energy stored in the compressed spring.
The potential energy (PE) stored in a spring can be calculated using the formula:
PE = (1/2) k x^2
Where k is the spring constant and x is the displacement from the relaxed state.
In this case, the spring constant, k, is given as 7.5 N/cm, or 0.075 N/mm.
The displacement, x, is given as 3.2 cm, which is equivalent to 0.032 m.
Substituting the values into the formula, we can calculate the potential energy stored in the spring:
PE = (1/2) (0.075 N/mm) (0.032 m)^2
= (1/2) (0.075 N/mm) (0.001024 m^2)
= 3.84 x 10^-5 Nm
Next, let's consider the kinetic energy (KE) of the ball as it leaves the barrel.
The kinetic energy of an object can be calculated using the formula:
KE = (1/2) m v^2
Where m is the mass of the ball and v is its velocity.
In this case, the mass of the ball, m, is given as 12 g, which is equivalent to 0.012 kg.
We want to find the velocity v at which the ball leaves the barrel.
Since the spring force does work on the ball, converting potential energy into kinetic energy, we can use the principle of conservation of mechanical energy. This states that the total mechanical energy before the ball is launched (potential energy) is equal to the total mechanical energy after the ball is launched (kinetic energy).
Therefore, we have:
PE = KE
Substituting the values, we get:
3.84 x 10^-5 Nm = (1/2) (0.012 kg) v^2
Simplifying the equation, we can solve for v:
v^2 = (2 * 3.84 x 10^-5 Nm) / (0.012 kg)
v^2 = 6.4 x 10^-3 m^2/s^2
Taking the square root of both sides, we find:
v = √(6.4 x 10^-3 m^2/s^2)
= 0.08 m/s
Therefore, the ball will leave the barrel with a speed of 0.08 m/s.