A spring whose spring constant is 650 N/m is compressed 0.300 m. What speed (in meters/second) can it give to a 0.300 kg ball when released
(1/2)kX^2 = (1/2)MV^2
Solve for V
thanks
To find the speed that the spring can give to the ball when released, we can use the law of conservation of energy. The potential energy stored in the compressed spring is converted into the kinetic energy of the ball when released.
The potential energy stored in a spring is given by the equation:
Potential Energy = 0.5 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring constant (k) is 650 N/m and the displacement (x) is 0.300 m. Substituting these values into the equation, we have:
Potential Energy = 0.5 * 650 N/m * (0.300 m)^2
Potential Energy = 29.25 J
According to the law of conservation of energy, this potential energy is converted into kinetic energy when the ball is released:
Kinetic Energy = 0.5 * m * v^2
where m is the mass of the ball (0.300 kg) and v is its velocity.
Equating the potential energy and the kinetic energy, we have:
29.25 J = 0.5 * 0.300 kg * v^2
Simplifying the equation, we get:
v^2 = (2 * 29.25 J) / 0.300 kg
v^2 = 195 J/kg
v = √(195 J/kg)
Using a calculator, we find:
v ≈ 13.94 m/s (rounded to two decimal places)
Therefore, the speed that the spring can give to the ball when released is approximately 13.94 meters per second.