Final answer:
The dart is initially moving with a velocity of 0.502 m/s.
Step-by-step explanation:
To calculate the initial velocity of the dart, we can use the principle of conservation of mechanical energy. When the dart is released, the potential energy stored in the spring is converted into kinetic energy. The potential energy stored in the spring is given by the equation:
PE = (1/2) kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.
In this case, the spring constant is 56 N/m and the displacement is 0.03 m. Plugging these values into the equation, we get:
PE = (1/2) * 56 * (0.03)^2 = 0.0252 J
Since the potential energy is converted into kinetic energy, we can set the potential energy equal to the kinetic energy:
KE = (1/2) mv^2
where KE is the kinetic energy, m is the mass of the dart, and v is the initial velocity of the dart.
In this case, the mass of the dart is 0.2 kg. Plugging in the potential energy and mass into the equation, we get:
0.0252 = (1/2) * 0.2 * v^2
Simplifying the equation, we find:
v^2 = 0.0252 * 2 / 0.2 = 0.252
Taking the square root of both sides, we get:
v = sqrt(0.252) = 0.502 m/s
Therefore, the dart is initially moving with a velocity of 0.502 m/s.