Questions: You weigh 530 N. Your bathroom scale contains a light but very stiff ideal spring. When you stand at rest on the scale, the spring is compressed 1.80 cm. Your 180 N dog then gently jumps into your arms. Part A How much work is done by the spring as the two of you are brought to rest by friction? Express your answer to three significant figures and include the appropriate units. W= Units

Solution Steps

First, we need to find the spring constant \( k \) of the bathroom scale. We know that when you stand on the scale, the spring is compressed by 1.80 cm (0.018 m) and the force exerted by your weight is 530 N.

Using Hooke's Law: \[ F = kx \]

Rearranging to solve for \( k \): \[ k = \frac{F}{x} = \frac{530 \, \text{N}}{0.018 \, \text{m}} = 29444.4444 \, \text{N/m} \]

Next, we need to find the total weight when your 180 N dog jumps into your arms. The combined weight is: \[ F_{\text{total}} = 530 \, \text{N} + 180 \, \text{N} = 710 \, \text{N} \]

Using Hooke's Law again, we can find the new compression \( x_{\text{new}} \) of the spring with the combined weight: \[ x_{\text{new}} = \frac{F_{\text{total}}}{k} = \frac{710 \, \text{N}}{29444.4444 \, \text{N/m}} = 0.0241 \, \text{m} \]

The work done by the spring as it compresses from the initial compression to the new compression can be found using the work-energy principle for springs: \[ W = \frac{1}{2} k (x_{\text{new}}^2 - x_{\text{initial}}^2) \]

Substituting the values: \[ W = \frac{1}{2} \times 29444.4444 \, \text{N/m} \times (0.0241^2 - 0.018^2) \, \text{m}^2 \] \[ W = \frac{1}{2} \times 29444.4444 \times (0.00058081 - 0.000324) \] \[ W = \frac{1}{2} \times 29444.4444 \times 0.00025681 \] \[ W = 3.780 \, \text{J} \]

Final Answer