A spring (k=302 N/m) placed vertically on a surface has a 548 g ball placed on top of it. How far must the spring be compressed in order for the ball to be traveling at a speed of 2.8 m/s when it is located 438 cm above its initial position?
To find the distance the spring must be compressed, we need to use the principle of conservation of mechanical energy.
The total mechanical energy of the system consists of two parts: the potential energy stored in the spring and the gravitational potential energy of the ball.
The potential energy stored in the spring is given by the equation:
Potential Energy (PE) = (1/2) * k * x^2
Where k is the spring constant (302 N/m) and x is the distance the spring is compressed.
The gravitational potential energy of the ball is given by the equation:
Potential Energy (PE) = m * g * h
Where m is the mass of the ball (548 g = 0.548 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the initial position (438 cm = 4.38 m).
Since the total mechanical energy is conserved, we can set the potential energy stored in the spring equal to the gravitational potential energy of the ball:
(1/2) * k * x^2 = m * g * h
Substituting the known values into the equation:
(1/2) * (302 N/m) * x^2 = (0.548 kg) * (9.8 m/s^2) * (4.38 m)
Now we can solve for x:
x^2 = (2 * (0.548 kg) * (9.8 m/s^2) * (4.38 m)) / (302 N/m)
x^2 = 0.233037
x ≈ 0.482 m
Therefore, the spring must be compressed approximately 0.482 meters in order for the ball to be traveling at a speed of 2.8 m/s when it is located 438 cm above its initial position.