A particle of mass m kg is attached to one end A of a model spring OA of natural length L=0.81m and stiffness K N/m. The spring is hung vertically with the end O fixed and end A attached to the particle, which moves in a vertical line. The particle's mass, stiffness and natural length of the spring are related to the expression KL=2mg, where g is the magnitude of the acceleration due to gravity. It is observed that when the particle is at its highest point during the motion, its distance below O is 1/2L.
Using the law of conservation of mechanical energy, find the distance, in m, of the particle below O when it is at its lowest point during the motion.
thanks!
To find the distance of the particle below O when it is at its lowest point during the motion, we can start by analyzing the situation.
At the highest point, the particle has potential energy but no kinetic energy. Therefore, the total mechanical energy at this point is equal to the potential energy.
The potential energy of the particle is given by the equation:
Potential energy = mgh
Where m is the mass of the particle, g is the acceleration due to gravity, and h is the height.
From the given information, we know that when the particle is at its highest point, the distance below O is 1/2L. Therefore, the height h is equal to L/2.
Substituting the values into the equation, we have:
Potential energy = mgh
= mg(L/2)
We also know that the total mechanical energy at the highest point is equal to the potential energy:
Total mechanical energy = mgh
= mg(L/2)
Now, let's determine the total mechanical energy at the lowest point, where the particle has no potential energy but maximum kinetic energy.
At the lowest point, the particle has no potential energy but maximum kinetic energy. Therefore, the total mechanical energy at this point is equal to the kinetic energy.
The kinetic energy of the particle is given by the equation:
Kinetic energy = (1/2)mv^2
Where m is the mass of the particle and v is the velocity.
Since the particle moves in a vertical line, at the lowest point, all the potential energy gets converted into the kinetic energy.
Therefore, we have:
Total mechanical energy = Kinetic energy
= (1/2)mv^2
Now, we want to find the distance of the particle below O when it is at its lowest point. Let's call this distance D.
Since the particle is at its lowest point, the spring is fully stretched, and the restoring force of the spring is at its maximum. At this point, the force exerted by the spring is equal to the weight of the particle.
According to Hooke's Law, the force exerted by the spring is given by the equation:
Force = K * distance
Where K is the stiffness of the spring and distance is the displacement of the spring from its natural length.
From the given expression KL = 2mg, we can solve for K:
K = 2mg/L
At the lowest point, the spring is completely stretched, so the total force exerted by the spring is equal to the weight of the particle:
Force = mg
Equating the two forces, we have:
K * distance = mg
Substituting the value of K, we get:
(2mg/L) * distance = mg
Multiplying both sides by L, we have:
2mg * distance = mg * L
Canceling out the mass, we have:
2g * distance = g * L
Dividing both sides by g, we get:
2 * distance = L
Simplifying, we have:
distance = L/2
Therefore, the distance of the particle below O when it is at its lowest point during the motion is L/2, which is equal to 0.81/2 = 0.405 m.
To find the distance of the particle below O when it is at its lowest point, we can use the law of conservation of mechanical energy.
According to the law of conservation of mechanical energy, the total mechanical energy of a system remains constant if there are no external forces doing work on the system. In this case, the only force is the force of gravity acting on the particle.
At the highest point, the particle has potential energy but no kinetic energy. At the lowest point, the particle has kinetic energy but no potential energy.
Let's denote the distance of the particle below O when it is at its lowest point as x.
The potential energy at the highest point is given by the formula:
Potential Energy = mgh
where m is the mass of the particle, g is the acceleration due to gravity, and h is the distance of the particle below O at the highest point. We are given that h = 1/2L.
Therefore, Potential Energy = mg(1/2L) = (mgL)/2
The kinetic energy at the lowest point is given by the formula:
Kinetic Energy = (1/2)mv^2
where v is the velocity of the particle at the lowest point.
According to the law of conservation of mechanical energy:
Potential Energy at highest point = Kinetic Energy at lowest point
(mgL)/2 = (1/2)mv^2
Simplifying the equation, we can cancel out the factors of (1/2) and m:
gL = v^2
Now, let's consider the relationship given in the problem:
KL = 2mg
Substituting the given value of KL, which is 2mg, we get:
2mg = 2mg
Therefore, KL = 2mg is satisfied.
Now, let's find the expression for v using the relationship KL = 2mg:
KL = 2mg
K(0.81) = 2mg (substituting the value of L, which is 0.81m)
K = 2g (canceling out the factors of m)
Now, substituting this value of K in gL = v^2:
2g(0.81) = v^2
1.62g = v^2
Taking the square root of both sides, we get:
v = sqrt(1.62g)
Therefore, the velocity of the particle at the lowest point is v = sqrt(1.62g).
Now, let's find the distance x:
Kinetic Energy = (1/2)mv^2
The mass of the particle is m, and we have already found the value of v. Substituting these values, we get:
(1/2)m(sqrt(1.62g))^2 = (1/2)mv^2
Simplifying the equation, we get:
m(1.62g) = (1/2)mv^2
Canceling out the factors of m, we get:
1.62g = (1/2)v^2
Substituting the value of v^2, which is 1.62g:
1.62g = (1/2)(1.62g)
Simplifying further, we get:
1.62g = 0.81g
Therefore, the distance x below O when the particle is at its lowest point during the motion is 0.81 meters.
So, the particle is at a distance of 0.81 meters below O when it is at its lowest point.