On top of a spiral spring of the force constant 500Nm-1 is placed a mass of 5 x 10-3 kg. If the spring is compressed downwards by a length of 0.02m and then...

On top of a spiral spring of the force constant 500Nm-1 is placed a mass of 5 x 10-3 kg. If the spring is compressed downwards by a length of 0.02m and then released, calculate the height to which the mass is projected

Answer Details

The problem involves calculating the height to which a mass will be projected after being released from a compressed spring. The solution requires the use of conservation of energy principles. When the spring is compressed, it has potential energy stored in it, given by the formula: U = (1/2)kx^2 where U is the potential energy stored in the spring, k is the force constant of the spring, and x is the compression of the spring. In this case, the spring has a force constant of 500 N/m and is compressed by a length of 0.02 m. Therefore, the potential energy stored in the spring is: U = (1/2) * 500 N/m * (0.02 m)^2 = 0.1 J When the spring is released, this potential energy is converted into kinetic energy as the mass moves upwards. The kinetic energy of the mass can be calculated using the formula: K = (1/2)mv^2 where K is the kinetic energy of the mass, m is the mass of the object, and v is its velocity. At the highest point of its trajectory, the mass will have zero velocity and maximum potential energy (due to its height). Therefore, the kinetic energy at this point will be zero. Since energy is conserved, we can equate the initial potential energy of the spring to the potential energy of the mass at the highest point of its trajectory: U = mgh where g is the acceleration due to gravity, and h is the maximum height attained by the mass. Solving for h, we get: h = U / (mg) = 0.1 J / (0.005 kg * 9.81 m/s^2) = 2.04 m Therefore, the height to which the mass is projected is approximately 2 meters.