A vertical spring has one end attached to the ceiling and a 2.50-kg weight attac
ID: 1628620 • Letter: A
Question
A vertical spring has one end attached to the ceiling and a 2.50-kg weight attached to the other one. When the system is at rest, the spring is stretched by 20.0 cm. Now let the weight drop from a position in which the spring is already stretched by 5.0 cm. Use the conservation of energy law to find: a) how fast the weight is moving when it goes through the equilibrium position: b) How far down the weight will drop before starting to come back. What is the weight's acceleration when it's at the lowest position (give the magnitude and show the direction)?Explanation / Answer
At the point when the system is very rest, the spring is extended by 20cm.
We should decide the spring constant. At the point when the system is rest ,the spring's force is supporting the weight.
Spring’s force = k * d,
d = distance the spring is stretched (in meters)
Spring’s force = k * 0.2
Weight = 2.5 * 9.8 = 24.5 N
k * 0.2 = 24.5
k = 122.5 N/m
The equilibrium position is when the system is at rest. At this position, the potential and kinetic energy of the weight is 0 J.
Let’s determine the potential energy of the spring, when the system is at rest.
PE = ½ * 122.5 * 0.2^2 = 2.45 J
When the system is at rest, the total energy of the system is 2.45 J.
To return the weight to the position in which the spring is not deformed at all, the weight must be pushed 20 cm upward.
At this position, the potential energy of the spring is 0 J. At this position, the potential energy of the fish = 2.5 * 9.8 * 0.2 = 4.9 J
Total energy = 4.9 J
As the object drops 20cm, its potential energy decreases 4.9 J.
As the spring stretches 20cm, its potential energy increases 2.45 J.
Total energy at this position = 4.9 – 2.45 = 2.45 J
This is the kinetic energy of the object at this position.
½ * 2.5 * v^2 = 2.45
v = 1.4 m/s
This is the speed of the weight after it drops 20 cm. This is the equilibrium position of the spring.
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b)
The weight will continue dropping, until its kinetic energy is 0 J instead of 2.45 J. So, the potential energy of the spring will increase 2.45 J.
½ * 122.5 * d^2 = 2.45
d = 0.2 m
This is the distance the drops before starting to come back up!
Since this is harmonic motion, the weight will continue moving 20 cm above the equilibrium position and 20 cm below the equilibrium position!