Answer:
The mass of the object involved and the value of the gravitational acceleration
Explanation:
- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:

where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.
- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:

where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.
As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.
Answer:
Explanation:
We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.
By using energy conservation
KE₁ + U₁ = KE₂ + U₂
KE₁=Kinetic energy at location 1
U₁ =Potential energy at location 1
KE₂=Kinetic energy at location 2
U₂=Potential energy at location 2
Therefore, Raymond is thinking in a right way.
The particles can undergo small oscillations around x₂.
The given parameters;
- <em>initial energy of the particles = E₁</em>
- <em>final energy of the particles, E₂ = 0.33E₁</em>
The movement of the particles depends on the kinetic energy of the particles.
When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.
However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.
- The maximum position the particle can oscillate is x₅
- The half position the particles can oscillate is x₃
Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.
Thus, we can conclude that the particles can undergo small oscillations around x₂.
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Answer: a) Mr = 2.4×10^-4kg/s
V = 34.42m/a
b) E = 173J
Ø = 2693.1J
c) Er = 0.64J/s
Explanation: Please find the attached file for the solution
Work done = Force x Distance
Force = 10 lb = 44.5 N
Work Done = 44.5 N x 15 m
= 667.5 N-m