Question

suppose a 0.70-kg mass on a spring that has been compressed 0.26 m has elastic potential energy of 2.20 j. how much further must the spring be compressed to triple the elastic potential energy?

Answer 1

The spring must be further compressed by approximately
`\( 3.45 \, \text{m} \)`to triple the elastic potential energy.

The elastic potential energy
`(\(U\))` stored in a spring is given by the formula:

`\[ U = (1)/(2)kx^2 \]`

where:

-
`\( U \)` is the elastic potential energy,

-
`\( k \)` is the spring constant,

-
`\( x \)` is the displacement from the equilibrium position.

In this case, you want to find how much further the spring must be compressed to triple the elastic potential energy. Let
`\( x_1 \)` be the initial displacement (0.26 m), and
`\( x_2 \)` be the additional displacement.

The elastic potential energy when the spring is compressed by
`\( x_1 \)` is given by:

`\[ U_1 = (1)/(2)kx_1^2 \]`

The elastic potential energy when the spring is further compressed by
`\( x_2 \)` (to triple the energy) is given by:

`\[ U_2 = (1)/(2)k(x_1 + x_2)^2 \]`

Given that
`\( U_1 = 2.20 \, \text{J} \)` and you want to triple the energy
`(\( U_2 = 3 * 2.20 \))`, you can set up the equation:

`\[ 3 * 2.20 = (1)/(2)k(x_1 + x_2)^2 \]`

Now, substitute the values and solve for
`\( x_2 \)`:

`\[ 6.60 = (1)/(2)k(0.26 + x_2)^2 \]`

Multiply both sides by
`\( (2)/(k) \)`:

`\[ 13.20 = (0.26 + x_2)^2 \]`

`\[ 0.26 + x_2 = √(13.20) \]\[ x_2 = √(13.20) - 0.26 \]\[ x_2 \approx 3.45 \, \text{m} \]`

Answer 2

The distance the spring needs to compressed to in order to triple the elastic potential energy is 0.45 m

How to calculate the distance the spring need to be compressed to?

First, we shall calculate the spring constant. This is shown below:

• Compression (e) = 0.26 m
• Elastic potential energy (E) = 2.20 J
• Spring constant (K) =?

`K = (2E)/(e^2) \\\\K = (2\ *\ 2.2)/(0.26^2)\\\\K = 65.09\ N/m\\\\`

Now, we shall obtain the distance the spring needs to be compressed to in order to have the energy tripled. This is shown below:

• Initial elastic energy (E) = 2.20 J
• New elastic potential energy (PE) = 3E = 3 × 2.20 = 6.6 J
• Spring constant (K) = 65.09 N/m
• Compression (e) = ?

`e = \sqrt{(2PE)/(K)} \\\\e = \sqrt{(2\ *\ 6.6)/(65.09)} \\\\e = 0.45\ m`