A horizontal spring attached to a wall has a force constant of 800 N/m. A block of mass 1.10 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released.
(a) What objects constitute the system, and through what forces do they interact?
(b) What are the two points of interest?
(c) Find the energy stored in the spring when the mass is stretched 6.20 cm from equilibrium and again when the mass passes through equilibrium after being released from rest.
x = 6.20 cm J
x = 0 cm J
(d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value.
m/s
(e) What is the speed at the halfway point?
m/s
Why isn't it half the speed at equilibrium?
(a) The objects that constitute the system in this scenario are the horizontal spring attached to the wall and the block of mass 1.10 kg attached to the spring. The spring and block interact through the force exerted by the spring, known as the spring force.
(b) The two points of interest in this scenario are the equilibrium point and the halfway point. The equilibrium point is the position where the spring is neither stretched nor compressed, and the block is at rest. The halfway point is the position halfway between the equilibrium point and the maximum displacement of the block.
(c) To find the energy stored in the spring when the mass is stretched 6.20 cm from equilibrium, we can use the formula for the potential energy stored in a spring:
Potential energy = (1/2) * k * x^2
where k is the force constant of the spring and x is the displacement from the equilibrium point. Substituting the given values:
Potential energy = (1/2) * 800 N/m * (0.0620 m)^2
Potential energy = 1.57 J
When the mass passes through equilibrium after being released from rest, the potential energy stored in the spring is zero because the spring is neither stretched nor compressed.
(d) The conservation of energy equation for this situation can be written as:
Total mechanical energy at release = Total mechanical energy at equilibrium
At release, the only form of energy is potential energy stored in the spring, which is given as 1.57 J. At equilibrium, the only form of energy is kinetic energy, which can be written as:
Kinetic energy = (1/2) * m * v^2
where m is the mass of the block and v is the speed of the block at equilibrium. Setting the potential and kinetic energies equal to each other:
(1/2) * 800 N/m * (0.0620 m)^2 = (1/2) * 1.10 kg * v^2
v^2 = (800 N/m * (0.0620 m)^2) / 1.10 kg
v = sqrt((800 N/m * (0.0620 m)^2) / 1.10 kg)
v ≈ 1.72 m/s
Therefore, the speed of the mass as it passes equilibrium is approximately 1.72 m/s.
(e) The speed at the halfway point is not half the speed at equilibrium because the relationship between displacement and velocity in simple harmonic motion is not linear. As the block oscillates between the equilibrium point and the maximum displacement, its velocity is constantly changing. The speed at the halfway point is actually at its maximum, which occurs when the block passes through the equilibrium point.