A 0.53-kg mass is attached to a horizontal spring with k = 106 N/m. The mass slides across a frictionless surface. The spring is stretched 27 cm from equilibrium, and then the mass is released from rest.
(a) Find the mechanical energy of the system.
(b) Find the speed of the mass when it has moved 9 cm.
(c) Find the maximum speed of the mass.
a) ME= 1/2 k x^2
b) 1/2 m v^2=change in PE= 1/2 k (.27^2-.09^2)
In turns out that for b), instead of using
(.27^2 -0.09^2), we have to use (.27^2-0.18^2).
I think your solution tells us when the spring moves 18 cm, not 9 cm.
Thank you, though. This was really useful.
To find the answers to these questions, we need to apply the principles of energy conservation and the spring-mass system equations. Let's go through each question step by step.
(a) To find the mechanical energy of the system, we need to consider the potential energy and the kinetic energy of the mass. At the starting position, the potential energy stored in the spring is given by 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium. In this case, the displacement is 27 cm (or 0.27 m). So the potential energy is:
Potential energy = (1/2)(106 N/m)(0.27 m)^2 = 3.45 J
Since the mass is initially at rest, it has zero kinetic energy. Therefore, the total mechanical energy of the system is equal to the potential energy:
Mechanical energy = Potential energy = 3.45 J
(b) To find the speed of the mass when it has moved 9 cm (or 0.09 m), we can use the principle of conservation of mechanical energy. At any point along its path, the total mechanical energy remains constant. So, we can equate the initial mechanical energy to the final mechanical energy:
Initial mechanical energy = Final mechanical energy
The initial mechanical energy is the potential energy at the starting position, which we calculated to be 3.45 J. The final mechanical energy consists of the potential energy and the kinetic energy of the mass when it has moved 9 cm. Let's assume the speed at this position is v:
Final mechanical energy = Potential energy + Kinetic energy
Final mechanical energy = (1/2)(106 N/m)(0.09 m)^2 + (1/2)(0.53 kg)(v^2)
Since the mass is attached to a horizontal spring, the potential energy is given by 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium. So:
Final mechanical energy = (1/2)(106 N/m)(0.09 m)^2 + (1/2)(0.53 kg)(v^2)
Now, we can equate the initial and final mechanical energies:
3.45 J = (1/2)(106 N/m)(0.09 m)^2 + (1/2)(0.53 kg)(v^2)
Simplifying and solving for v^2, we get:
v^2 = (2(3.45 J) - (0.5)(106 N/m)(0.09 m)^2) / (0.53 kg)
v^2 ≈ 6.803
v ≈ 2.61 m/s
Therefore, when the mass has moved 9 cm, its speed is approximately 2.61 m/s.
(c) To find the maximum speed of the mass, we need to consider the point where all the potential energy has been converted into kinetic energy. This occurs at the maximum displacement from equilibrium, which is equal to the amplitude of the oscillation. In this case, the amplitude is 0.27 m. So, at this point:
Potential energy = 0
Kinetic energy = Total mechanical energy
Using the formula for kinetic energy, we have:
Total mechanical energy = (1/2)(0.53 kg)(v_max^2)
Since the potential energy is zero, the total mechanical energy is equal to the kinetic energy:
(1/2)(0.53 kg)(v_max^2) = 3.45 J
Simplifying and solving for v_max^2, we get:
v_max^2 = (2(3.45 J)) / (0.53 kg)
v_max^2 ≈ 12.962
v_max ≈ 3.60 m/s
Therefore, the maximum speed of the mass is approximately 3.60 m/s.