Question

Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage–Earth system equals twice the gain in the spring’s potential energy.

Short answer

The loss of gravitational potential energy of the cabbage-earth system equals twice the gain in spring’s potential energy, i.e., $U_g=2U_e$.

Step-by-step solution

Given data:

Gravitational force is equal to the restoring force of the spring.

To understand the concept:

Here spring force is balanced by weight. You can use this fact to find the spring constant. To find the loss of gravitational potential energy, you can use the formula for spring elastic potential energy.

Formula:

Write the equation for the spring force as below.

F = kx

Here, F is the spring force, k is the spring constant, and x is the displacement.

Step 3: The loss of gravitational potential energy of the cabbage–Earth system equals twice the gain in the spring’s potential energy:

Let x be the displacement of spring, so spring force is given as,

F = kx

This spring force is balanced by weight, so

kx = m x g

Here, m is the mass and g is the acceleration due to gravity.

$\Rightarrow k=m\frac{g}{x}$

Now, gain in spring potential energy is given as follows:

$U_e=\frac{1}{2}k{x}^2$

Now plug the value of k in the above equation.

role="math" localid="1661237598418" $$\begin{gathered} U_e=\frac{1}{2}\left(m\frac{g}{x}\right)x^2 \\ {U}_e=\frac{1}{2}mgx............\left(1\right) \end{gathered}$$

But the loss in gravitational potential energy is given as

$U_g=mgx$ ….. (2)

Comparing equations (1) and (2), you obtain

$U_e=\frac{1}{2}{U}_{g}U$

Therefore,

$U_g=2U_e$

Hence, the loss of gravitational potential energy of the cabbage-earth system equals twice the gain in spring’s potential energy is $U_g=2U_e$.