An object that is 0.5 m above the ground has the same amount of potential energy as a spring that is stretched 0.5 m. Each distance is then doubled.

Question

How will the potential energies of the object and the spring compare after the distances are doubled?

The gravitational potential energy of the object will be two times greater than the elastic potential energy of the spring.
The elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.
The elastic potential energy of the spring will be two times greater than the gravitational potential energy of the object.
The potential energies will remain equal to one another.

Answer 1

To compare the gravitational potential energy of the object with the elastic potential energy of the spring after the distances are doubled, we can use the formulas for each type of potential energy.

1. Gravitational Potential Energy (U): The formula is given by: $$U=mgh$$ where $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height above the ground.

2. Elastic Potential Energy (PE): The formula for the elastic potential energy stored in a spring is: $$PE=\frac{1}{2}k{x}^2$$ where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Initially, we have:

• Height $h=0.5,\text{m}$ for the object.
• Displacement $x=0.5,\text{m}$ for the spring.

After doubling the distances:

• Height will be $h^{\prime }=1.0,\text{m}$.
• Displacement will be $x^{\prime }=1.0,\text{m}$.

Now, let's evaluate the potential energies after the distances are doubled.

1. Gravitational Potential Energy after doubling height: $$U^{\prime }=mg(1.0)=2mg(0.5)=2U$$

2. Elastic Potential Energy after doubling the displacement: $$P{E}^{\prime }=\frac{1}{2}k(1.0{)}^2=\frac{1}{2}k(0.5{)}^2\cdot 4=4PE$$

Initially, we found that the gravitational potential energy of the object was equal to the elastic potential energy of the spring at the half meter stretch, thus we had: $$U=PE$$

After doubling:

• $U^{\prime }$ becomes $2U$ (or 2 times the original gravitational potential energy).
• $P{E}^{\prime }$ becomes $4PE$ (or 4 times the original spring potential energy).

Since initially $U=PE$, after doubling the distances, we have:

• $U^{\prime }=2PE$
• $P{E}^{\prime }=4PE$

Thus, the elastic potential energy of the spring will become four times greater than the gravitational potential energy of the object.

Therefore, the correct statement is: The elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.