Question

A 0.290 kg frame, when suspended from a coil spring, stretches the spring 0.0400 mm. A 0.200 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm.

Find the maximum distance the frame moves downward from its initial equilibrium position?
I got d= 0.1286 m, but it's wrong.
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Answer 1

The maximum distance the frame moves downward from its initial equilibrium position is
`\(0.0312 \, \text{m}\)`.

To find the maximum distance the frame moves downward after the lump of putty is dropped onto it, we can use the principle of conservation of mechanical energy.

The potential energy of the lump of putty just before it hits the frame will be converted into the potential energy of the frame-putty system at its maximum downward displacement.

The potential energy of the lump of putty just before it hits the frame can be calculated using the formula:

`\[ PE_{\text{lump}} = mgh \]`

Where:

`\(m\) =` mass of the putty = 0.200 kg

`\(g\) =` acceleration due to gravity = 9.81 m/s²

`\(h\) =` height from which the putty is dropped = 30.0 cm = 0.30 m

`\[ PE_{\text{lump}} = 0.200 \, \text{kg} * 9.81 \, \text{m/s}^2 * 0.30 \, \text{m} = 0.5886 \, \text{J} \]`

Now, at the maximum downward displacement, this potential energy will be converted into the potential energy of the frame-putty system, considering the spring has been stretched by the added mass of the putty.

The potential energy stored in the spring is given by:

`\[ PE_{\text{spring}} = (1)/(2)kx^2 \]`

Where:

`\(k\) =` spring constant

`\(x\) =` displacement from equilibrium position

First, let's calculate the spring constant k using Hooke's Law:

`\[ F_{\text{spring}} = -kx \]`

The force exerted by the spring can be equated to the weight of the frame and the putty at the maximum displacement:

`\[ F_{\text{spring}} = mg \]`

Where:

`\(m\)` = mass of the frame + mass of the putty = 0.290 kg + 0.200 kg = 0.490 kg

`\(g\) =` acceleration due to gravity = 9.81 m/s²

`\[ kx = mg \]`

`\[ k = (mg)/(x) \]`

`\[ k = \frac{0.490 \, \text{kg} * 9.81 \, \text{m/s}^2}{0.0400 * 10^(-3) \, \text{m}} \]`

`\[ k = 1209.225 \, \text{N/m} \]`

Now, equate the potential energies:

`\[ PE_{\text{lump}} = PE_{\text{spring}} \]`

`\[ 0.5886 \, \text{J} = (1)/(2) * 1209.225 \, \text{N/m} * x^2 \]`

`\[ x^2 = \frac{2 * 0.5886 \, \text{J}}{1209.225 \, \text{N/m}} \]`

`\[ x^2 = 9.712 * 10^(-4) \, \text{m}^2 \]`

Taking the square root:

`\[ x = \sqrt{9.712 * 10^(-4) \, \text{m}^2} \]`

`\[ x = 0.0312 \, \text{m} \]`

Therefore, The answer is
`\(0.0312 \, \text{m}\)`.

Answer 2

The maximum distance the frame moves downward from its initial equilibrium position is 4.07 mm.

How to calculate the maximum distance?

The maximum distance the frame moves downward from its initial equilibrium position is calculated as follows;

The spring constant of the spring;

F = kx

mg = kx

k = mg/x

k = (0.29 kg x 9.8 m/s² ) / (0.04 x 10⁻³ m)

k = 71,050 N/m

The extension of the spring when the putty is dropped on it;

U = P.E

¹/₂kx² = mgh

x² = (2mgh/k)

x = √ (2mgh/k)

x = √ ( 2 x 0.2 x 9.8 x 0.3 / 71,050)

x = 4.07 x 10⁻³ m

x = 4.07 mm