A physics student wants to build a spring gun that when aimed horizontally will give a 15.0- g ball an initial acceleration of 'g' by releasing a spring that has been compressed 15.0 cm. The spring constant is .981. What is the velocity of the ball as it emerges from the gun?
To find the velocity of the ball as it emerges from the gun, you can use the principle of conservation of energy. The potential energy stored in the compressed spring will be converted into the kinetic energy of the ball as it is released.
First, let's find the potential energy stored in the compressed spring. The formula for potential energy stored in a spring is given by:
Potential energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this case, the spring has been compressed by 15.0 cm, which is equal to 0.15 m. The spring constant is given as 0.981 N/m. Plugging these values into the formula, we get:
PE = (1/2) * 0.981 N/m * (0.15 m)^2
Next, we'll equate the potential energy to the kinetic energy of the ball. The formula for kinetic energy is given by:
Kinetic energy (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 is given as 15.0 g, which is equal to 0.015 kg. We want to find the velocity of the ball when it emerges from the gun, so we can rewrite the equation as:
PE = KE
(1/2) * 0.981 N/m * (0.15 m)^2 = (1/2) * 0.015 kg * v^2
Now, solve the equation for v^2:
v^2 = (0.981 N/m * (0.15 m)^2) / 0.015 kg
v^2 = (0.981 N/m * 0.0225 m^2) / 0.015 kg
v^2 = 14.715 N*m / 0.015 kg
v^2 = 981 m^2/s^2
Finally, take the square root of both sides to find the velocity:
v = √(981 m^2/s^2)
v ≈ 31.3 m/s
Therefore, the velocity of the ball as it emerges from the gun is approximately 31.3 m/s.