A 2.00kg mass is hanging on a vertical compressed spring 1.0 m above the floor.The spring constant is 75N/m and the spring is compressed 0.90m. The mass is released so that it falls vertically.Determine
1) The acceleration of the mass when it was dropped.
2) The speed of the spring/mass when it is 0.6m above the floor
To find the answers to these questions, we can apply the principles of energy conservation and Newton's laws of motion.
1) To determine the acceleration of the mass when it was dropped, we need to consider the forces acting on it. When the mass is released, the only force acting on it is gravity. Initially, the spring is compressed, so it does not contribute to the forces acting on the mass.
The force due to gravity can be calculated using the equation F = mg, where m is the mass (2.00 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Thus, the force due to gravity is Fg = 2.00 kg * 9.8 m/s^2 = 19.6 N.
Since the spring is initially compressed, it exerts a force in the opposite direction. According to Hooke's Law, the force exerted by the spring is given by Fs = -kx, where k is the spring constant (75 N/m) and x is the displacement from the equilibrium position (0.90 m).
The net force acting on the mass is given by the sum of the force due to gravity and the force exerted by the spring (Fnet = Fg + Fs). Since the forces act in opposite directions, we have Fnet = Fg - Fs.
Substituting the values, we get Fnet = 19.6 N - (-75 N/m * 0.90 m) = 19.6 N + 67.5 N = 87.1 N.
According to Newton's second law, F = ma, where F is the net force and m is the mass. So, we can calculate the acceleration (a) by rearranging the equation to a = F/m. Substituting the values, we find a = 87.1 N / 2.00 kg = 43.55 m/s^2.
Therefore, the acceleration of the mass when it was dropped is approximately 43.55 m/s^2.
2) To find the speed of the spring/mass when it is 0.6 m above the floor, we can again apply energy conservation principles.
First, let's find the potential energy at that position. The potential energy of the spring mass system is given by the equation U = (1/2) k x^2, where k is the spring constant (75 N/m) and x is the displacement from the equilibrium position (0.90 m - 0.6 m = 0.30 m).
Substituting the values, we get U = (1/2) * 75 N/m * (0.30 m)^2 = 3.375 J.
The potential energy at this position is converted into kinetic energy when the mass is in motion. So, we can equate the potential energy to the kinetic energy to find the speed.
The kinetic energy is given by the equation KE = (1/2) m v^2, where m is the mass (2.00 kg) and v is the speed.
Setting the potential energy equal to the kinetic energy, we have U = KE.
Substituting the values, we get 3.375 J = (1/2) * 2.00 kg * v^2.
Simplifying the equation, we find v^2 = (2 * 3.375 J) / (2.00 kg) = 3.375 m^2/s^2.
Taking the square root of both sides, we find v = √3.375 m/s ≈ 1.838 m/s.
Therefore, the speed of the spring/mass when it is 0.6 m above the floor is approximately 1.838 m/s.